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

10. Find the equation of the circle passing through the points (4,1) and (6,5) and whose centre is on the line 4x + y = 16.

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Payal Gupta | Contributor-Level 10

4 months ago

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 – 12

...more

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 = ${(\sqrt10)}^{2}$

(x – 3)² + (y – 4)² = 10

x² $+$ $y^{2} - 6 x - 8 y + 15 = 0$

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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)242 + h2 – 2.4.h + 12 + k2 – 2.1.k = 62 + h2 – 2.6.h + 52 + k2 – 2.5.k16 + h2 – 8h + 1 + k2 – 2k = 36 + h2 – 12h + 25 + k2 – 10k17 + h2 – 8h + k2 – 2k = 61 + h2  – 12h + k2 – 10k–8h + 12h – 2k + 10k = 61 – 174h + 8k = 444 (h + 2k) = 44h + 2k = <math><mrow><mfrac><mrow><mn>4</mn><mn>4</mn></mrow><mrow><mn>4</mn></mrow></mfrac></mrow></math> = 11- (iv)Since the centre of the circle passes through 4x + y = 16? 4h + k = 16 (4h + k = 16) × 28h + 2k = 32- (v)Subtract (v) – (iv).8h + 2k = 32– (h + 2k = 11)7h = 21h = <math><mrow><mfrac><mrow><mn>2</mn><mn>1</mn></mrow><mrow><mn>7</mn></mrow></mfrac></mrow></math> = 3Putting h = 3 in (iv), we get3 + 2k = 112k = 11 – 3k = <math><mrow><mfrac><mrow><mn>8</mn></mrow><mrow><mn>2</mn></mrow></mfrac></mrow></math> = 4Putting the values of h and k in, (ii) (4 – 3)2 + (1 – 4)2 = r212 + (–3)2 = r21 + 9 = r2r = √10<math><mrow></mrow></math>Thus the equation of the circle is, (x – 3)2 + (y – 4)2 = <math><mrow><msup><mrow><mrow><mo> (√10</mo><mrow></mrow><mo>)</mo></mrow></mrow><mrow><mn>2</mn></mrow></msup></mrow></math> (x – 3)2 + (y – 4)2 = 10x2<math><mo></mo><mi></mi><mrow><mo>+</mo></mrow><mprescripts></mprescripts></math><math><mrow><msup><mrow><mi>y</mi></mrow><mrow><mn>2</mn></mrow></msup><mo>−</mo><mn>6</mn><mi>x</mi><mo>−</mo><mn>8</mn><mi>y</mi><mo>+</mo><mn>1</mn><mn>5</mn><mo>=</mo><mn>0</mn></mrow></math>

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10. Find the equation of the circle passing through the points (4,1) and (6,5) and whose centre is on the line 4x + y = 16.

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