Page 33 - Exemplar Solution - Updates, Questions & Discussions on Exemplar Solution at Shiksha

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

Find the equation of the hyperbola with:

Vertices $(\pm 5, 0)$ , foci $(\pm 7, 0)$ .

Vertices $(0, \pm 7)$ , eccentricity $e = \sqrt[]{2}$ .

Foci $(0, \pm 10)$ , passing through (2, 3).

8 Views|Posted a year ago

Asked by Alok Kumar Singh

S

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} (i) G i v e n t h a t v e r t i c e s (\pm 5, 0), f o c i (\pm 7, 0) \\ V e r t e x o f h y p e r b o l a = (\pm a, 0) a n d f o c i (\pm a e, 0) \\ ∴ a = 5 a n d a e = 7 \Rightarrow 5 \times e = 7 \Rightarrow e = \frac{7}{5} \\ N o w b^{2} = a^{2} (e^{2} - 1) \\ \Rightarrow b^{2} = 25 (\frac{49}{25} - 1) \Rightarrow b^{2} = 25 \times \frac{24}{25} \Rightarrow b^{2} = 24 \\ T h e e q u a t i o n o f t h e h y p e r b o l a i s \\ \frac{x^{2}}{25} - \frac{y^{2}}{24} = 1 \\ (i i) G i v e n t h a t v e r t i c e s (0, \pm 7), e = \frac{4}{3} \\ C l e a r l y, t h e h y p e r b o l a i s v e r t i c a l . \\ ∴ a = 5 a n d e = \frac{4}{3} \\ W e k n o w t h a t b^{2} = a^{2} (e^{2} - 1) \\ \Rightarrow b^{2} = 49 (\frac{16}{9} - 1) \Rightarrow b^{2} = 49 \times \frac{7}{9} \Rightarrow b^{2} = \frac{343}{9} \\ T h e e q u a t i o n o f t h e h y p e r b o l a i s \\ \frac{y^{2}}{49} - \frac{9 x^{2}}{343} = 1 \Rightarrow 9 x^{2} - 7 y^{2} + 343 = 0 \\ (i i i) G i v e n t h a t : f o c i (0, \pm \sqrt[]{10}) \\ ∴ a e = \sqrt[]{10} \Rightarrow a^{2} e^{2} = 10 \\ W e k n o w t h a t b^{2} = a^{2} (e^{2} - 1) \\ \Rightarrow b^{2} = a^{2} e^{2} - a^{2} \Rightarrow b^{2} = 10 - a^{2} \\ E q u a t i o n o f t h e h y p e r b o l a i s \\ \frac{y^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} = 1 \Rightarrow \frac{y^{2}}{a^{2}} - \frac{x^{2}}{10 - a^{2}} = 1 \\ I f i t p a s s e s t h r o u g h t h e (2, 3) t h e n \\ \Rightarrow \frac{9}{a^{2}} - \frac{4}{10 - a^{2}} = 1 \Rightarrow \frac{90 - 9 a^{2} - 4 a^{2}}{a^{2} (10 - a^{2})} = 1 \\ \Rightarrow 90 - 13 a^{2} = 10 a^{2} - a^{4} \Rightarrow a^{4} - 23 a^{2} + 90 = 0 \\ \Rightarrow a^{4} - 18 a^{2} - 5 a^{2} + 90 = 0 \\ \Rightarrow a^{2} (a^{2} - 18) - 5 (a^{2} - 18) = 0 \\ \Rightarrow (a^{2} - 18) (a^{2} - 5) = 0 \\ \Rightarrow a^{2} = 18, a^{2} = 5 \\ ∴ b^{2} = 10 - 18 = - 8 a n d b^{2} = 10 - 5 = 5 \\ ∴ b \neq - 8 a n d b^{2} = 5 \\ H e n c e, t h e r e q u i r e d e q u a t i o n i s \frac{y^{2}}{5} - \frac{x^{2}}{5} = 1 o r y^{2} - x^{2} = 5 . \end{array}$

Upvote

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

Show that the set of all points such that the difference of their distances from (4, 0) and (-4, 0) is always equal to 2 represents a hyperbola.

5 Views|Posted a year ago

Asked by Alok Kumar Singh

S

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t (x, y) b e a n ypoint \\ A c c o r d i n g t o t h e q u e s t i o n, w e h a v e \\ \sqrt[]{{(x - 4)}^{2} + {(y - 0)}^{2}} - \sqrt[]{{(x + 4)}^{2} + {(y - 0)}^{2}} = 2 \\ \Rightarrow \sqrt[]{x^{2} + 16 - 8 x + y^{2}} - \sqrt[]{x^{2} + 16 + 8 x + y^{2}} = 2 \\ P u t t i n g x^{2} + y^{2} + 16 = z \dots (i) \\ \Rightarrow \sqrt[]{z - 8 x} - \sqrt[]{z + 8 x} = 2 \\ S q u a r i n g b o t h s i d e s, w e h a v e \\ \Rightarrow z - 8 x + z + 8 x - 2 \sqrt[]{(z - 8 x) (z + 8 x)} = 4 \\ \Rightarrow 2 z - 2 \sqrt[]{z^{2} - 64 x^{2}} = 4 \\ \Rightarrow z - \sqrt[]{z^{2} - 64 x^{2}} = 2 \\ \Rightarrow (z - 2) = \sqrt[]{z^{2} - 64 x^{2}} \\ A g a i n s q u a r i n g b o t h s i d e s, w e g e t \\ \Rightarrow z^{2} + 4 - 4 z = z^{2} - 64 x^{2} \\ \Rightarrow 4 - 4 z + 64 x^{2} = 0 \\ P u t t i n g t h e v a l u e o f z, w e h a v e \\ \Rightarrow 4 - 4 (x^{2} + y^{2} + 16) + 64 x^{2} = 0 \\ \Rightarrow 4 - 4 x^{2} - 4 y^{2} - 64 + 64 x^{2} = 0 \\ \Rightarrow 60 x^{2} - 4 y^{2} - 60 = 0 \\ \Rightarrow 60 x^{2} - 4 y^{2} = 60 \\ \Rightarrow \frac{60 x^{2}}{60} - \frac{4 y^{2}}{60} = 1 \\ \Rightarrow \frac{x^{2}}{1} - \frac{y^{2}}{15} = 1 \\ W h i c h r e p r e s e n t a h p e r b o l a . H e n c e p r o v e d . \end{array}$

Upvote

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

Find the equation of the set of all points whose distance from (0, 4) equals their distance from the line $y = 9$

4 Views|Posted a year ago

Asked by Alok Kumar Singh

S

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t (x, y) b e a n y \\ A c c o r d i n g t o t h e q u e s t i o n, w e h a v e \\ \sqrt[]{{(x - 0)}^{2} + {(y - 4)}^{2}} = \frac{2}{3} | \frac{y - 9}{1} | \\ S q u a r i n g b o t h s i d e s, w e h a v e \\ \Rightarrow x^{2} + {(y - 4)}^{2} = \frac{4}{9} (y^{2} + 81 - 18 y) \\ \Rightarrow 9 x^{2} + 9 {(y - 4)}^{2} = 4 y^{2} + 324 - 72 y \\ \Rightarrow 9 x^{2} + 9 y^{2} + 144 - 72 y = 4 y^{2} + 324 - 72 y \\ \Rightarrow 9 x^{2} + 5 y^{2} + 144 - 324 = 0 \\ \Rightarrow 9 x^{2} + 5 y^{2} - 180 = 0 \\ H e n c e, t h e r e q u i r e d e q u a t i o n i s 9 x^{2} + 5 y^{2} - 180 = 0 . \end{array}$

Upvote

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

Find the equation of the set of all points the sum of whose distances from the points (3, 0) and (9, 0) is 12.

2 Views|Posted a year ago

Asked by Alok Kumar Singh

S

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t (x, y) b e a n y \\ G i v e n a r e (3, 0) a n d (9, 0) \\ A c c o r d i n g t o t h e q u e s t i o n, w e h a v e \\ \sqrt[]{{(x - 3)}^{2} + {(y - 0)}^{2}} + \sqrt[]{{(x - 9)}^{2} + {(y - 0)}^{2}} = 12 \\ \Rightarrow \sqrt[]{x^{2} + 9 - 6 x + y^{2}} + \sqrt[]{x^{2} + 81 - 18 x + y^{2}} = 12 \\ P u t t i n g x^{2} + 9 - 6 x + y^{2} = k \\ \Rightarrow \sqrt[]{k} + \sqrt[]{72 - 12 x + k} = 12 \\ \Rightarrow \sqrt[]{72 - 12 x + k} = 12 - \sqrt[]{k} \\ S q u a r i n g b o t h s i d e s, w e h a v e \\ \Rightarrow 72 - 12 x + k = 144 + k - 24 \sqrt[]{k} \\ \Rightarrow 24 \sqrt[]{k} = 144 - 72 + 12 x \\ \Rightarrow 24 \sqrt[]{k} = 72 + 12 x \\ \Rightarrow 2 \sqrt[]{k} = 6 + x \\ A g a i n s q u a r i n g b o t h s i d e s, w e g e t \\ \Rightarrow 4 k = 36 + x^{2} + 12 x \\ P u t t i n g t h e v a l u e o f k, w e h a v e \\ 4 (x^{2} + 9 - 6 x + y^{2}) = 36 + x^{2} + 12 x \\ \Rightarrow 4 x^{2} + 36 - 24 x + 4 y^{2} = 36 + x^{2} + 12 x \\ \Rightarrow 3 x^{2} + 4 y^{2} - 36 x = 0 \\ H e n c e, t h e r e q u i r e d e q u a t i o n i s 3 x^{2} + 4 y^{2} - 36 x = 0 . \end{array}$

Upvote

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

Find the equation of each of the following parabolas:

Directrix $x = 0$ , focus at (6, 0).

Vertex at (0, 4), focus at (0, 2).

Focus at (-1,-2), directrix $x - 2 y + 3 = 0$ .

3 Views|Posted a year ago

Asked by Alok Kumar Singh

S

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} (i) G i v e n t h a t d i r e c t r i x = 0 a n d f o c u s (6, 0) \\ ∴ T h e e q u a t i o n o f t h e p a r a b o l a i s \\ {(x - 6)}^{2} + y^{2} = x^{2} \\ \Rightarrow x^{2} + 36 - 12 x + y^{2} = x^{2} \\ \Rightarrow y^{2} - 12 x + 36 = 0 \\ H e n c e, t h e r e q u i r e d e q u a t i o n i s y^{2} - 12 x + 36 = 0 \\ (i i) G i v e n t h a t v e r t e x a t (0, 4) a n d f o c u s a t (0, 2) . \\ S o, e q u a t i o n o f d i r e c t r i x i s y - 6 = 0 \\ A c c o r d i n g t o t h e d e f i n i t i o n o f t h e p a r a b o l a \\ P F = P M \\ \sqrt[]{{(x - 0)}^{2} + {(y - 2)}^{2}} = | y - 6 | \\ \Rightarrow \sqrt[]{x^{2} + y^{2} + 4 - 4 y} = | y - 6 | \\ S q u a r i n g b o t h s i d e s, w e g e t \\ x^{2} + y^{2} + 4 - 4 y = y^{2} + 36 - 12 y \\ \Rightarrow x^{2} + 4 - 4 y = 36 - 12 y \\ \Rightarrow x^{2} + 8 y - 32 = 0 \\ \Rightarrow x^{2} = 32 - 8 y \\ H e n c e, t h e r e q u i r e d e q u a t i o n i s x^{2} = 32 - 8 y \\ (i i i) G i v e n t h a t f o c u s a t (- 1, - 2) a n d d i r e c t r i x x - 2 y + 3 = 0 \\ L e t (x, y) b e a n y o n t h e p a r a b o l a . \\ A c c o r d i n g t o t h e d e f i n i t i o n o f t h e p a r a b o l a \\ P F = P M \\ \sqrt[]{{(x + 1)}^{2} + {(y + 2)}^{2}} = | \frac{x - 2 y + 3}{\sqrt[]{{(1)}^{2} + {(- 2)}^{2}}} | \\ \Rightarrow \sqrt[]{x^{2} + 1 + 2 x + y^{2} + 4 + 4 y} = | \frac{x - 2 y + 3}{\sqrt[]{5}} | \\ S q u a r i n g b o t h s i d e s, w e g e t \\ x^{2} + 1 + 2 x + y^{2} + 4 + 4 y = \frac{x^{2} + 4 y^{2} + 9 - 4 x y - 12 y + 6 x}{5} \\ \Rightarrow 5 x^{2} + 5 + 10 x + 5 y^{2} + 20 + 20 y = x^{2} + 4 y^{2} + 9 - 4 x y - 12 y + 6 x \\ \Rightarrow 4 x^{2} + y^{2} + 4 x y + 4 x + 32 y + 16 = 0 \\ H e n c e, t h e r e q u i r e d e q u a t i o n i s 4 x^{2} + y^{2} + 4 x y + 4 x + 32 y + 16 = 0 . \end{array}$

Upvote

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

Find the equation of a circle passing through the point (7, 3), having a radius of 3 units, and whose center lies on the line $y = x - 1$ .

10 Views|Posted a year ago

Asked by Alok Kumar Singh

S

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t t h e e q u a t i o n o f t h e c i r c l e b e \\ {(x - h)}^{2} + {(y - k)}^{2} = r^{2} \\ I f i t p a s s e s t h r o u g h (7, 3) t h e n \\ {(7 - h)}^{2} + {(3 - k)}^{2} = {(3)}^{2} [? r = 3] \\ \Rightarrow 49 + h^{2} - 14 h + 9 + k^{2} - 6 k = 9 \\ \Rightarrow h^{2} + k^{2} - 14 h - 6 k + 49 = 0 \dots (i) \\ I f c e n t r e (h, k) l i e s o n t h e l i n e y = x - 1 t h e n \\ k = h - 1 \dots (i i) \\ P u t t i n g t h e v a l u e o f k i n e q n . (i) w e g e t \\ h^{2} + {(h - 1)}^{2} - 14 h - 6 (h - 1) + 49 = 0 \\ \Rightarrow h^{2} + h^{2} + 1 - 2 h - 14 h - 6 h + 6 + 49 = 0 \\ \Rightarrow 2 h^{2} - 22 h + 56 = 0 \\ \Rightarrow 2 (h^{2} - 11 h + 28) = 0 \\ \Rightarrow h^{2} - 11 h + 28 = 0 \\ \Rightarrow h^{2} - 7 h - 4 h + 28 = 0 \\ \Rightarrow h (h - 7) - 4 (h - 7) = 0 \\ \Rightarrow (h - 7) (h - 4) = 0 \Rightarrow h = 7, 4 \\ F r o m e q n . (i i) w e g e t k = 4 - 1 = 3 a n d k = 7 - 1 = 6 . \\ S o, t h e c e n t r e s a r e (4, 3) a n d (7, 6) . \\ E q u a t i o n o f t h e c i r c l e i s \\ T a k i n g c e n t r e (4, 3) \\ {(x - 4)}^{2} + {(y - 3)}^{2} = 9 \\ \Rightarrow x^{2} + 16 - 8 x + y^{2} + 9 - 6 y = 9 \\ \Rightarrow x^{2} + y^{2} - 8 x - 6 y + 16 = 0 \\ T a k i n g c e n t r e (7, 6) \\ {(x - 7)}^{2} + {(y - 6)}^{2} = 9 \\ \Rightarrow x^{2} + 49 - 14 x + y^{2} + 36 - 12 y = 9 \\ \Rightarrow x^{2} + y^{2} - 14 x - 12 y + 76 = 0 \\ H e n c e, t h e r e q u i r e d e q u a t i o n s a r e \\ x^{2} + y^{2} - 8 x - 6 y + 16 = 0 \\ a n d x^{2} + y^{2} - 14 x - 12 y + 76 = 0 \end{array}$

Upvote

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

Find the equation of a circle of radius 5 which is touching another circle $x^{2} + y^{2} - 2 x - 4 y - 20 = 0$ at (5, 5).

4 Views|Posted a year ago

Asked by Alok Kumar Singh

S

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n c i r c l e i s = (3, - 1) \\ x^{2} + y^{2} - 2 x - 4 y - 20 = 0 \\ 2 g = - 2 \Rightarrow g = - 1 \\ 2 f = - 4 \Rightarrow f = - 2 \\ ∴ C e n t r e C_{1} = (1, 2) \\ a n d r a d i u s r = \sqrt[]{g^{2} + f^{2} - c} \\ = \sqrt[]{1 + 4 + 20} = 5 \\ L e t t h e c e n t r e o f t h e r e q u i r e d c i r c l e b e (h, k) . \\ C l e a r l y, P i s t h e m i d o f C_{1} C_{2} \\ ∴ 5 = \frac{1 + h}{2} \Rightarrow h = 9 \\ a n d 5 = \frac{2 + k}{2} \Rightarrow k = 8 \\ R a d i u s o f t h e r e q u i r e d c i r c l e = 5 \\ ∴ E q u a t i o n o f t h e c i r c l e i s \\ {(x - 9)}^{2} + {(y - 8)}^{2} = {(5)}^{2} \\ \Rightarrow x^{2} + 81 - 18 x + y^{2} + 64 - 16 y = 25 \\ \Rightarrow x^{2} + y^{2} - 18 x - 16 y + 145 - 25 = 0 \\ \Rightarrow x^{2} + y^{2} - 18 x - 16 y + 120 = 0 \\ H e n c e, t h e r e q u i r e d e q u a t i o n i s x^{2} + y^{2} - 18 x - 16 y + 120 = 0 . \end{array}$

Upvote

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

Find the equation of a circle whose center is (3, -1) and which cuts off a chord of length 6 units on the line $2 x - 5 y + 18 = 0$ .

[Hint: To determine the radius of the circle, find the perpendicular distance from the center to the given line.]

5 Views|Posted a year ago

Asked by Alok Kumar Singh

S

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : \\ C e n t r e o f t h e c i r c l e = (3, - 1) \\ L e n g t h o f c h o r d A B = 6 u n i t s \\ C P = | \frac{2 \times 3 - 5 \times 1 + 18}{\sqrt[]{{(2)}^{2} + {(- 5)}^{2}}} | \\ = | \frac{6 + 5 + 18}{\sqrt[]{29}} | = \sqrt[]{29} \\ N o w A B = 6 u n i t s . \\ ∴ A P = \frac{1}{2} A B = \frac{1}{2} \times 6 = 3 u n i t s \\ I n Δ C P A, A C^{2} = C P^{2} + A P^{2} \\ = {(\sqrt[]{29})}^{2} + {(3)}^{2} = 29 + 9 = 38 \\ ∴ A C = \sqrt[]{38} \\ S o, t h e r a d i u s o f t h e c i r c l e, r = \sqrt[]{38} \\ ∴ E q u a t i o n o f t h e c i r c l e i s \\ {(x - 3)}^{2} + {(y + 1)}^{2} = {(\sqrt[]{38})}^{2} \\ \Rightarrow {(x - 3)}^{2} + {(y + 1)}^{2} = 38 \\ \Rightarrow x^{2} + 9 - 6 x + y^{2} + 1 + 2 y = 38 \\ \Rightarrow x^{2} + y^{2} - 6 x + 2 y = 28 \\ H e n c e, t h e r e q u i r e d e q u a t i o n i s x^{2} + y^{2} - 6 x + 2 y = 28 . \end{array}$

Upvote

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

Find the equation of the circle which passes through the points (2, 3) and (4, 5), and its center lies on the straight line $y - 4 x + 3 = 0$ .

3 Views|Posted a year ago

Asked by Alok Kumar Singh

S

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t t h e e q u a t i o n o f t h e c i r c l e i s \\ {(x - h)}^{2} + {(y - k)}^{2} = r^{2} \dots (i) \\ I f t h e c i r c l e passes t h r o u g h (2, 3) a n d (4, 5) t h e n \\ {(2 - h)}^{2} + {(3 - k)}^{2} = r^{2} \dots (i i) \\ a n d {(4 - h)}^{2} + {(5 - k)}^{2} = r^{2} \dots (i i i) \\ S u b t r a c t i n g e q n . (i i i) f r o m e q n . (i i) w e h a v e \\ {(2 - h)}^{2} - {(4 - h)}^{2} + {(3 - k)}^{2} - {(5 - k)}^{2} = 0 \\ \Rightarrow 4 + h^{2} - 4 h - 16 - h^{2} + 8 h + 9 + k^{2} - 6 k - 25 - k^{2} + 10 k = 0 \\ \Rightarrow 4 h + 4 k - 28 = 0 \\ \Rightarrow h + k = 7 \dots (i v) \\ , t h e c e n t r e (h, k) l i e s o n t h e l i n e y - 4 x + 3 = 0 \\ t h e n k - 4 h - 3 = 0 \Rightarrow k = 4 h - 3 \\ P u t t i n g t h e v a l u e o f k i n e q n . (i v) w e g e t \\ h + 4 h - 3 = 7 \Rightarrow 5 h = 10 \Rightarrow h = 2 \\ F r o m (i v) w e g e t k = 5 \\ P u t t i n g t h e v a l u e o f h a n d k i n e q n . (i i) w e g e t \\ {(2 - 2)}^{2} + {(3 - 5)}^{2} = r^{2} \Rightarrow r^{2} = 4 \\ S o, t h e e q u a t i o n o f t h e c i r c l e i s \\ {(x - 2)}^{2} + {(y - 5)}^{2} = 4 \\ \Rightarrow x^{2} + 4 - 4 x + y^{2} + 25 - 10 y = 4 \\ \Rightarrow x^{2} + y^{2} - 4 x - 10 y + 25 = 0 \\ H e n c e, t h e r e q u i r e d e q u a t i o n i s x^{2} + y^{2} - 4 x - 10 y + 25 = 0 . \end{array}$

Upvote

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

If the lines $2 x - 3 y = 5$ and $3 x - 4 y = 7$ are the diameters of a circle with an area of 154 square units, then obtain the equation of the circle.

13 Views|Posted a year ago

Asked by Alok Kumar Singh

S

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} W e k n o w t h a t t h e intersection point o f t h e d i a m e t e r g i v e s t h e c e n t r e o f t h e c i r c l e . \\ G i v e n e q u a t i o n s o f d i a m e t e r s a r e \\ 2 x - 3 y = 5 \dots (i) \\ 3 x - 4 y = 7 \dots (i i) \\ F r o m e q n . (i) w e h a v e x = \frac{5 + 3 y}{2} \dots (i i i) \\ P u t t i n g t h e v a l u e o f x i n e q n . (i i) w e h a v e \\ 3 (\frac{5 + 3 y}{2}) - 4 y = 7 \\ \Rightarrow 15 + 9 y - 8 y = 14 \Rightarrow y = 14 - 15 \Rightarrow y = - 1 \\ N o w f r o m e q n . (i i i) w e h a v e x = \frac{5 + 3 (- 1)}{2} \Rightarrow x = \frac{5 - 3}{2} \Rightarrow x = 1 \\ S o, t h e c e n t r e o f t h e c i r c l e = (1, - 1) \\ G i v e n t h a t a r e a o f t h e c i r c l e = 154 \\ \Rightarrow π r^{2} = 154 \\ \Rightarrow \frac{22}{7} \times r^{2} = 154 \Rightarrow r^{2} = 154 \times \frac{7}{22} \\ \Rightarrow r^{2} = 7 \times 7 \Rightarrow r = 7 \\ S o, t h e e q u a t i o n o f t h e c i r c l e i s \\ {(x - 1)}^{2} + {(y + 1)}^{2} = {(7)}^{2} \\ \Rightarrow x^{2} + 1 - 2 x + y^{2} + 1 + 2 y = 49 \\ \Rightarrow x^{2} + y^{2} - 2 x + 2 y = 47 \\ H e n c e, t h e r e q u i r e d e q u a t i o n o f t h e c i r c l e i s x^{2} + y^{2} - 2 x + 2 y = 47 . \end{array}$

Upvote

Exemplar Solution

Questions

Discussions

Active Users

Followers

Find the equation of the hyperbola with:

Vertices $(\pm 5, 0)$ , foci $(\pm 7, 0)$ .

Vertices $(0, \pm 7)$ , eccentricity $e = \sqrt[]{2}$ .

Foci $(0, \pm 10)$ , passing through (2, 3).

Show that the set of all points such that the difference of their distances from (4, 0) and (-4, 0) is always equal to 2 represents a hyperbola.

Find the equation of the set of all points whose distance from (0, 4) equals their distance from the line $y = 9$

Find the equation of the set of all points the sum of whose distances from the points (3, 0) and (9, 0) is 12.

Find the equation of each of the following parabolas:

Directrix $x = 0$ , focus at (6, 0).

Vertex at (0, 4), focus at (0, 2).

Focus at (-1,-2), directrix $x - 2 y + 3 = 0$ .

Find the equation of a circle passing through the point (7, 3), having a radius of 3 units, and whose center lies on the line $y = x - 1$ .

Find the equation of a circle of radius 5 which is touching another circle $x^{2} + y^{2} - 2 x - 4 y - 20 = 0$ at (5, 5).

Find the equation of a circle whose center is (3, -1) and which cuts off a chord of length 6 units on the line $2 x - 5 y + 18 = 0$ .

[Hint: To determine the radius of the circle, find the perpendicular distance from the center to the given line.]

Find the equation of the circle which passes through the points (2, 3) and (4, 5), and its center lies on the straight line $y - 4 x + 3 = 0$ .

If the lines $2 x - 3 y = 5$ and $3 x - 4 y = 7$ are the diameters of a circle with an area of 154 square units, then obtain the equation of the circle.