Ncert Solutions Maths class 11th

13. Let the equation of the circle be,

(x – h)² + (y – k)² = r² – 0-----(i)

As the circle is passing through (0, 0) we get,

(0 – h)² + (0 – k)² = r²

(–h)² + (–k)² = r²

h² + k² = r²--------(ii)

The circle also makes intercepts a and b on the cooperate axes.

Let intercept on x–axis be 'a' and on y–axis be 'b'

?Circle also passes through (a, 0) and (0, b)

Putting x = a and y = 0 in (i)

(a – h)² + (0 – k)² = r²

a² + h² – 2.a.h + (–k)² = r²

a² – 2ah + h² + k² = r²

Putting value of r² from (ii), we get

a² – 2ah + h² + k² = h² + k²

a² = 2ah

a = 2h

h = $\frac{a}{2}$

Putting x = 0 andy = b in (i).

(0 – h)² + (b – k)² = r²

h² + b² + k² – 2bk = r²

Putting value of r² from (ii),

h² + b² + k² – 2bk = h² + k²

b² = 2bk

b = 2k

k = $\frac{b}{2}$

Putting values of h and k in (ii),

${\left(\frac{a}{2}\right)}^2+{\left(\frac{b}{2}\right)}^2=r^2$

$\frac{a^2}{4}+\frac{b^2}{4}=r^2$

r² = $\frac{a^2+b^2}{4}$

Putting the values of h₁ k and r² in (i),

${\left(x-\frac{a}{2}\right)}^2+{\left(y-\frac{b}{2}\right)}^2=\frac{a^2+b^2}{4}$

x² + $\frac{a^2}{4}$ – 2.x. $\frac{a}{2}$ + y² + ${\left(\frac{b}{2}\right)}^2$ – 2 .y. $\frac{b}{2}$ = $\frac{a^2+b^2}{4}$

x² + $\cancel{\frac{a^2}{4}}$ – ax + y² + $\cancel{\frac{b^2}{4}}$ – yb = $\cancel{\frac{a^2}{4}}+\cancel{\frac{b^2}{4}}$

?The equation of the circle is,

x² – ax + y² – by = 0

12. Given,

r = 5.

Since the centre lies on x – axis.

k = 0.

? Centre of circle = (h, k) = (h, 0)

The equation of the circle in given by,

(x – h)² + (y – k)² = r²

(x – h)² + (y – 0)² = (5)²

(x – h)² + y² = 25- (i)

Since the curie parses through the point (2, 3),

Putting x = 2 and y = 3 in equation (1), We get.

(2 – h)² + (3)² = 25

2² + h² – 22.h + 9 = 25

4 + h²– 4h + 9 = 25

h² – 4h + 13 – 25 = 0

h² – 4h – 12 = 0

h² – (6 – 2)h – 12 = 0

h² – 6h + 2h – 12 – 0

h (h – 6) + 2 (h – 6) = 0

(h – 6) (h + 2) = 0

h = 6 and h = –2

When h = 6,

From (i), Equation of circle is,

(x –6)² + y² = 25

(x – 6)² + (y – 0)² = 52

When h = –2,

Equation of circle is, (x + 2)² + (y – 0)² = 5² or x² +2$y^2+4x-21=0$$$

11. Let the equation of the circle be.

(x – h)² + (y – k)² = r² -----(i)

Since the circle passes through (2, 3) and (–1, 1),

Putting x = 2 and y = 3 in (i),

(2 – h)² + (3 – k)² = r²---------(ii)

Putting x = –1 and y = 1 in (i),

(–1 – h)² + (1 – k)² = r²

Equating equation (ii) and (iii), We get:–

(2 – h)² + (3 – k)² = (–1 – h)² + (1 – k)²

2² + h² – 2.2.h + 3² + k² – 2.3.k = (–1)² + h² – 2.(–1).h + 1² + k² – 2(1)(k)

4 + h² – 4h + 9 + k² – 6k = 1 + h² + 2h + 1 + k² – 2k

13 – 4h – 6k = 2 + 2h – 2k

–4h – 2h – 6k + 2k = 2 – 13

–6h – 4k = 11

–(6h + 4k) = 11

6h + 4k = 11-------(iv)

Since the centre of the circle is on the line x – 3y – 11 = 0

? h – 3k = 11

(h – 3k = 11) * 6 = 6h – 18k = 66-----(v)

Subtract (v) – (iv),

$\begin{array}{l}\cancel{6h}-18k=66\\ \underset{\_}{\left(\cancel{6h}\pm 4k=-11\right)}\end{array}$

– 22k = 55

k = $\frac{-55}{22}=\frac{-5}{22}$

Putting the value of k in (iv), we get

6h + $\cancel{\stackrel{\mathrm{2/4}}{}}$ * $\left(-\frac{55}{\underset{11}{\cancel{22}}}\right)=11$

6h + $\left(-\frac{110}{11}\right)=11$

6h = 11 + $\frac{\cancel{\stackrel{10}{110}}}{\underset{1}{\cancel{11}}}$

6h = 11 + 10

h = $\frac{{\cancel{21}}^7}{\cancel{\underset{2}{6}}}=\frac{7}{2}$

Putting values of h and k in (ii), we get

${\left(2-\frac{7}{2}\right)}^2+{\left(3+\frac{5}{2}\right)}^2=r^2$

${\left(\frac{4-7}{2}\right)}^2+{\left(\frac{6+5}{2}\right)}^2=r^2$

${\left(\frac{-3}{2}\right)}^2+{\left(\frac{11}{2}\right)}^2=r^2$

$\frac{9}{4}+\frac{121}{4}=r^2$

$\frac{9+121}{4}=r^2$

r² = $\frac{\stackrel{65}{\cancel{130}}}{\cancel{\underset{2}{4}}}$

r² = $\frac{65}{2}$

Than the equation of the circle is,

${\left(x-\frac{7}{2}\right)}^2+{\left(y+\frac{5}{2}\right)}^2=\frac{65}{2}.$

$\Rightarrow x{}_2+y^2-7x+5y-14=0$

10. Let the equation of the circle be,

(x – h)² + (y – k)² = r² - (i)

Since the circle passes through (4, 1) and (6, 5)

Putting x = 4 and y = 1 in (i),

(4 – h)² + (1 – k)² = 2² - (ii)

Putting x = 6 and y = 5 in (i),

(6 – h)² + (5 – k)² = r²- (iii)

Equating equation (ii) and (iii), We get.

(4 – h)² + (1 – k)² = (6 – h)² + (5 – k)²

4² + h² – 2.4.h + 1² + k² – 2.1.k = 6² + h² – 2.6.h + 5² + k² – 2.5.k

16 + h² – 8h + 1 + k² – 2k = 36 + h² – 12h + 25 + k² – 10k

17 + h² – 8h + k² – 2k = 61 + h²  – 12h + k² – 10k

–8h + 12h – 2k + 10k = 61 – 17

4h + 8k = 44

4 (h + 2k) = 44

h + 2k = $\frac{44}{4}$ = 11- (iv)

Since the centre of the circle passes through 4x + y = 16

? 4h + k = 16

(4h + k = 16) * 2

8h + 2k = 32- (v)

Subtract (v) – (iv).

8h + 2k = 32

– (h + 2k = 11)

7h = 21

h = $\frac{21}{7}$ = 3

Putting h = 3 in (iv), we get

3 + 2k = 11

2k = 11 – 3

k = $\frac{8}{2}$ = 4

Putting the values of h and k in, (ii)

(4 – 3)² + (1 – 4)² = r²

1² + (–3)² = r²

1 + 9 = r²

r = √10$$

Thus the equation of the circle is,

(x – 3)² + (y – 4)² = ${(\surd 10)}^2$

(x – 3)² + (y – 4)² = 10

x²$+$$y^2-6x-8y+15=0$