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Conic Sections Ncert Solutions Maths class 11th Maths Class 11th Conic Sections

11. Find the equation of the circle passing through the points (2,3) and (–1,1) and whose centre is on the line x – 3y – 11 = 0.

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Payal Gupta | Contributor-Level 10

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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 –

...more

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),

$\begin{array}{l} 6 h - 18 k = 66 \\ \underline{(6 h \pm 4 k = - 11)} \end{array}$

– 22k = 55

k = $\frac{- 55}{22} = \frac{- 5}{22}$

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

6h + $\overset{2/4}{}$ × $(- \frac{55}{\underset{11}{22}}) = 11$

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

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

6h = 11 + 10

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

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

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

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

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

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

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

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

r² = $\frac{65}{2}$

Than the equation of the circle is,

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

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

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⇒ $\frac{{(\frac{24}{10})}^{2}}{a^{2}} + \frac{{(\frac{32}{10})}^{2}}{b^{2}} = 1$

...more

Slope of axis = $\frac{1}{2}$

$y - 3 = \frac{1}{2} (x - 2)$

⇒ 2y – 6 = x – 2

⇒ 2y – x – 4 = 0

2x + y – 6 = 0

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α + 1.6 = 4 ⇒ α = 2.4

β + 2.8 = 6 ⇒ β = 3.2

Ellipse passes through (2.4, 3.2)

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11. Find the equation of the circle passing through the points (2,3) and (–1,1) and whose centre is on the line x – 3y – 11 = 0.

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