Maths NCERT Exemplar Solutions Class 11th Chapter Eleven: Overview, Questions, Preparation

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

This is a Long Answer Type Questions as classified in NCERT Exemplar Sol: We know that the intersection point of the diameter gives the centre of the circle.Given equations of diameters are 2x−3y=5 …(i) 3x−4y=7 …(ii)From eqn.(i) we have x=5+3y2 …(iii)Putting the value of x in eqn.(ii) we have 3(5+3y2)−4y=7⇒ 15+9y−8y=14 ⇒y=14−15 ⇒y=−1Now from eqn.(iii) we have x=5+3(−1)2 ⇒x=5−32 ⇒x=1So, the centre of the circle=(1,−1)Given that area of the circle=154⇒ πr2=154⇒227×r2=154 ⇒r2=154×722⇒ r2=7×7 ⇒r=7So, the equation of the circle is (x−1)2+(y+1)2=(7)2⇒x2+1−2x+y2+1+2y=49⇒ x2+y2−2x+2y=47Hence, the required equation of the circle is x2+y2−2x+2y=47.

Q: Find the equation of a circle whose center is (3, -1) and which cuts off a chord of length 6 units on the line 2x−5y+18=0 . [Hint: To determine the radius of the circle, find the perpendicular distance from the center to the given line.]

This is a Long Answer Type Questions as classified in NCERT Exemplar Sol: Given that: Centre of the circle=(3,−1)Length of chord AB=6 units CP=|2×3−5×1+18(2)2+(−5)2| =|6+5+1829|=29Now AB=6 units.∴ AP=12AB=12×6=3 unitsIn ΔCPA, AC2=CP2+AP2 =(29)2+(3)2=29+9=38∴ AC=38So, the radius of the circle, r=38∴ Equation of the circle is (x−3)2+(y+1)2=(38)2⇒ (x−3)2+(y+1)2=38⇒x2+9−6x+y2+1+2y=38⇒ x2+y2−6x+2y=28Hence, the required equation is x2+y2−6x+2y=28.

Q: Find the equation of a circle of radius 5 which is touching another circle x2+y2−2x−4y−20=0 at (5, 5).

This is a Long Answer Type Questions as classified in NCERT Exemplar Sol: Given circle is=(3,−1)x2+y2−2x−4y−20=0 2g=−2 ⇒g=−1 2f=−4 ⇒f=−2∴ Centre C1=(1,2)and radius r=g2+f2−c =1+4+20=5Let the centre of the required circle be (h,k).Clearly, P is the mid of C1C2∴ 5=1+h2 ⇒h=9and 5=2+k2 ⇒k=8Radius of the required circle=5∴ Equation of the circle is (x−9)2+(y−8)2=(5)2⇒ x2+81−18x+y2+64−16y=25⇒x2+y2−18x−16y+145−25=0⇒ x2+y2−18x−16y+120=0Hence, the required equation is x2+y2−18x−16y+120=0.

Q: Find the equation of the circle which touches both axes in the first quadrant and whose radius is a .

This is a short answer type question as classified in NCERT Exemplar C l e a r l y c e n t r e o f t h e c i r c l e = ( a , a ) a n d r a d i u s = a E q u a t i o n o f c i r c l e w i t h r a d i u s r a n d c e n t r e ( h , k ) i s ( x − h ) 2 + ( y − k ) 2 = r 2 S o , t h e e q u a t i o n o f r e q u i r e d c i r c l e ⇒ ( x − a ) 2 + ( y − a ) 2 = a 2 ⇒ x 2 − 2 a x + a 2 + y 2 − 2 a y + a 2 = a 2 ⇒ x 2 + y 2 − 2 a x − 2 a y + a 2 = 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 − 2 a x − 2 a y + a 2 = 0 .

Q: Show that the point ( x , y ) given by x = 2 a t 1 + t 2 and y = a ( 1 − t 2 ) 1 + t 2 lies on a circle for all real values of t , such that − 1 < t < 1 , where a is any given real number.

This is a short answer type question as classified in NCERT Exemplar G i v e n t h a t : x = 2 a t 1 + t 2 a n d y = a ( 1 − t 2 ) 1 + t 2 ⇒ x 2 + y 2 = ( 2 a t 1 + t 2 ) 2 + ( a ( 1 − t 2 ) 1 + t 2 ) 2 = 4 a 2 t 2 ( 1 + t 2 ) 2 + a 2 ( 1 − t 2 ) 2 ( 1 + t 2 ) 2 = 4 a 2 t 2 + a 2 ( 1 + t 4 − 2 t 2 ) ( 1 + t 2 ) 2 = 4 a 2 t 2 + a 2 + a 2 t 4 − 2 a 2 t 2 ( 1 + t 2 ) 2 = a 2 + a 2 t 4 + 2 a 2 t 2 ( 1 + t 2 ) 2 = a 2 ( 1 + t 4 + 2 t 2 ) ( 1 + t 2 ) 2 = a 2 ( 1 + t 2 ) 2 ( 1 + t 2 ) 2 = a 2 ∴ x 2 + y 2 = a 2 w h i c h i s t h e e q u a t i o n o f a c i r c l e . Hence, the given points lie on a circle.

Updated on Sep 17, 2025 14:52 IST

By Payal Gupta, Retainer

Table of contents

Conic Section Short Answer Type Questions
Conic Section Long Answer Type Questions
Conic Section True or False Type Questions
Conic Section Fill in the Blanks Type Questions
Conic Section Objective Type Questions
JEE_Mains_01 MT

Conic Section Short Answer Type Questions

1. Find the equation of the circle which touches both axes in the first quadrant and whose radius is $a$ .

Sol.

\begin{array}{l} C l e a r l y c e n t r e o f t h e c i r c l e = (a, a) \\ a n d r a d i u s = a \\ E q u a t i o n o f c i r c l e w i t h r a d i u s r a n d c e n t r e (h, k) i s \\ {(x - h)}^{2} + {(y - k)}^{2} = r^{2} \\ S o, t h e e q u a t i o n o f r e q u i r e d c i r c l e \\ \Rightarrow {(x - a)}^{2} + {(y - a)}^{2} = a^{2} \\ \Rightarrow x^{2} - 2 a x + a^{2} + y^{2} - 2 a y + a^{2} = a^{2} \\ \Rightarrow x^{2} + y^{2} - 2 a x - 2 a y + a^{2} = 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} - 2 a x - 2 a y + a^{2} = 0 . \end{array}

2. Show that the point $(x, y)$ given by $x = \frac{2 a t}{1 + t^{2}}$ and $y = \frac{a (1 - t^{2})}{1 + t^{2}}$ lies on a circle for all real values of $t$ , such that $- 1 < t < 1$ , where $a$ is any given real number.

Sol.

\begin{array}{l} G i v e n t h a t : x = \frac{2 a t}{1 + t^{2}} a n d y = \frac{a (1 - t^{2})}{1 + t^{2}} \\ \Rightarrow x^{2} + y^{2} = {(\frac{2 a t}{1 + t^{2}})}^{2} + {(\frac{a (1 - t^{2})}{1 + t^{2}})}^{2} \\ = \frac{4 a^{2} t^{2}}{{(1 + t^{2})}^{2}} + \frac{a^{2} {(1 - t^{2})}^{2}}{{(1 + t^{2})}^{2}} = \frac{4 a^{2} t^{2} + a^{2} (1 + t^{4} - 2 t^{2})}{{(1 + t^{2})}^{2}} \\ = \frac{4 a^{2} t^{2} + a^{2} + a^{2} t^{4} - 2 a^{2} t^{2}}{{(1 + t^{2})}^{2}} = \frac{a^{2} + a^{2} t^{4} + 2 a^{2} t^{2}}{{(1 + t^{2})}^{2}} \\ = \frac{a^{2} (1 + t^{4} + 2 t^{2})}{{(1 + t^{2})}^{2}} = \frac{a^{2} {(1 + t^{2})}^{2}}{{(1 + t^{2})}^{2}} = a^{2} \\ ∴ x^{2} + y^{2} = a^{2} w h i c h i s t h e e q u a t i o n o f a c i r c l e . \\ H e n c e, t h e g i v e n points l i e o n a c i r c l e . \end{array}

Commonly asked questions

Find the equation of the circle which touches both axes in the first quadrant and whose radius is $a$ .

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} C l e a r l y c e n t r e o f t h e c i r c l e = (a, a) \\ a n d r a d i u s = a \\ E q u a t i o n o f c i r c l e w i t h r a d i u s r a n d c e n t r e (h, k) i s \\ {(x - h)}^{2} + {(y - k)}^{2} = r^{2} \\ S o, t h e e q u a t i o n o f r e q u i r e d c i r c l e \\ \Rightarrow {(x - a)}^{2} + {(y - a)}^{2} = a^{2} \\ \Rightarrow x^{2} - 2 a x + a^{2} + y^{2} - 2 a y + a^{2} = a^{2} \\ \Rightarrow x^{2} + y^{2} - 2 a x - 2 a y + a^{2} = 0 \\ H e n c e, t h e r e q u i r e d e q u a t i o n i s x^{2} + y^{2} - 2 a x - 2 a y + a^{2} = 0 . \end{array}$

Show that the point $(x, y)$ given by $x = \frac{2 a t}{1 + t^{2}}$ and $y = \frac{a (1 - t^{2})}{1 + t^{2}}$ lies on a circle for all real values of $t$ , such that $- 1 < t < 1$ , where $a$ is any given real number.

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : x = \frac{2 a t}{1 + t^{2}} a n d y = \frac{a (1 - t^{2})}{1 + t^{2}} \\ \Rightarrow x^{2} + y^{2} = {(\frac{2 a t}{1 + t^{2}})}^{2} + {(\frac{a (1 - t^{2})}{1 + t^{2}})}^{2} \\ = \frac{4 a^{2} t^{2}}{{(1 + t^{2})}^{2}} + \frac{a^{2} {(1 - t^{2})}^{2}}{{(1 + t^{2})}^{2}} = \frac{4 a^{2} t^{2} + a^{2} (1 + t^{4} - 2 t^{2})}{{(1 + t^{2})}^{2}} \\ = \frac{4 a^{2} t^{2} + a^{2} + a^{2} t^{4} - 2 a^{2} t^{2}}{{(1 + t^{2})}^{2}} = \frac{a^{2} + a^{2} t^{4} + 2 a^{2} t^{2}}{{(1 + t^{2})}^{2}} \\ = \frac{a^{2} (1 + t^{4} + 2 t^{2})}{{(1 + t^{2})}^{2}} = \frac{a^{2} {(1 + t^{2})}^{2}}{{(1 + t^{2})}^{2}} = a^{2} \\ ∴ x^{2} + y^{2} = a^{2} w h i c h i s t h e e q u a t i o n o f a c i r c l e . \\ H e n c e, t h e g i v e n points l i e o n a c i r c l e . \end{array}$

If a circle passes through the points $(0, 0), (a, 0), (0, b)$ , then find the coordinates of its center.

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n points a r e (0, 0), (a, 0) a n d (0, b) \\ G e n e r a l e q u a t i o n o f t h e c i r c l e i s \\ x^{2} + y^{2} + 2 g x + 2 f y + c = 0 w h e r e t h e c e n t r e i s (- g, - f) a n d r a d i u s = \sqrt[]{g^{2} + f^{2} - c} \\ I f i t p a s s e s t h r o u g h (0, 0) \\ ∴ c = 0 \\ I f i t p a s s e s t h r o u g h (a, 0) a n d (0, b) t h e n \\ a^{2} + 2 g a + c = 0 \Rightarrow a^{2} + 2 g a = 0 [? c = 0] \\ ∴ g = - \frac{a}{2} \\ a n d 0 + b^{2} + 0 + 2 f b + c = 0 \Rightarrow b^{2} + 2 f b = 0 [? c = 0] \\ \Rightarrow f = - \frac{b}{2} \\ H e n c e, t h e c o o r d i n a t e s o f c e n t r e o f t h e c i r c l e a r e (- g, - f) = (\frac{a}{2}, \frac{b}{2}) \end{array}$

Find the equation of the circle which touches the $x$ -axis and whose center is $(1, 2)$ .

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} S i n c e t h e c i r c l e w h o s e c e n t r e i s (1, 2) t o u c h x - a x i s \\ ∴ r = 2 \\ S o, t h e e q u a t i o n o f t h e c i r c l e i s \\ {(x - h)}^{2} + {(y - k)}^{2} = r^{2} \\ \Rightarrow {(x - 1)}^{2} + {(y - 2)}^{2} = {(2)}^{2} \\ \Rightarrow x^{2} - 2 x + 1 + y^{2} - 4 y + 4 = 4 \\ \Rightarrow x^{2} + y^{2} - 2 x - 4 y + 1 = 0 \\ H e n c e, t h e r e q u i r e d e q u a t i o n i s x^{2} + y^{2} - 2 x - 4 y + 1 = 0 . \end{array}$

If the lines $3 x - 4 y + 4 = 0$ and $6 x - 8 y - 7 = 0$ are tangents to a circle, then find the radius of the circle.
[Hint: Distance between given parallel lines gives the diameter of the circle.]