The circle passing through the intersection of the circles, <math><msup><mrow><mrow><mi>x</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup><mo>+</mo><msup><mrow><mrow><mi>y</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup><mo>-</mo><mn>6</mn><mi>x</mi><mo>=</mo><mn>0</mn></math> and <math><msup><mrow><mrow><mi>x</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup><mo>+</mo><msup><mrow><mrow><mi>y</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup><mo>-</mo><mn>4</mn><mi>y</mi></math> <math><mo>=</mo><mn>0</mn></math> , having its centre on the line, <math><mn>2</mn><mi>x</mi><mo>-</mo><mn>3</mn><mi>y</mi><mo>+</mo><mn>12</mn><mo>=</mo><mn>0</mn></math> , also passes through the point:

(a+2bcos?x)(a-2bcos?y)=a2-b2⇒a2-2abcos?y+2abcos?x-2b2cos?xcos?y=a2-b2Differentiating both sides:0-2ab-sin?ydydx+2ab(-sin?x)-2b2cos?x-sin?ydydx+cos?y(-sin?x)=0At π4,π4 :abdydx-ab-2b2-12dydx-12=0⇒dxdy=ab+b2ab-b2=a+ba-b;a,b>0

The circle passing through the intersection of the circles,&nbsp;<math><msup><mrow><mrow><mi>x</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup><mo>+</mo><msup><mrow><mrow><mi>y</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup><mo>-</mo><mn>6</mn><mi>x</mi><mo>=</mo><mn>0</mn></math>&nbsp;and&nbsp;<math><msup><mrow><mrow><mi>x</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup><mo>+</mo><msup><mrow><mrow><mi>y</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup><mo>-</mo><mn>4</mn><mi>y</mi></math>&nbsp;<math><mo>=</mo><mn>0</mn></math>&nbsp;, having its centre on the line,&nbsp;<math><mn>2</mn><mi>x</mi><mo>-</mo><mn>3</mn><mi>y</mi><mo>+</mo><mn>12</mn><mo>=</mo><mn>0</mn></math>&nbsp;, also passes through the point:

<math> <mo>(</mo> <mn>1</mn> <mo>,</mo> <mo>-</mo> <mn>3</mn> <mo>)</mo> </math> (Circles)

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Maths NCERT Exemplar Solutions Class 12th Chapter Twelve Maths Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Ten

The circle passing through the intersection of the circles, $x^{2} + y^{2} - 6 x = 0$ and $x^{2} + y^{2} - 4 y$ $= 0$ , having its centre on the line, $2 x - 3 y + 12 = 0$ , also passes through the point:

Option 1 -

$(- 1,3)$

Option 2 -

$(- 3,6)$

Option 3 -

$(- 3,1)$

Option 4 -

$(1, - 3)$ (Circles)

2 Views | Posted 2 months ago

Asked by Shiksha User

1 Answer

Answered by

alok kumar singh | Contributor-Level 10

2 months ago

Correct Option - 3

Detailed Solution:

$(a + \sqrt[]{2} b c o s ? x) (a - \sqrt[]{2} b c o s ? y) = a^{2} - b^{2}$

$\Rightarrow a^{2} - \sqrt[]{2} a b c o s ? y + \sqrt[]{2} a b c o s ? x - 2 b^{2} c o s ? x c o s ? y = a^{2} - b^{2}$

Differentiating both sides:

$0 - \sqrt[]{2} a b (- s i n ? y \frac{d y}{d x}) + \sqrt[]{2} a b (- s i n ? x) - 2 b^{2} [c o s ? x (- s i n ? y \frac{d y}{d x}) + c o s ? y (- s i n ? x)] = 0$

At $(\frac{π}{4}, \frac{π}{4})$ :

$\begin{array}{r} a b \frac{d y}{d x} - a b - 2 b^{2} (- \frac{1}{2} \frac{d y}{d x} - \frac{1}{2}) = 0 \\ \Rightarrow \frac{d x}{d y} = \frac{a b + b^{2}}{a b - b^{2}} = \frac{a + b}{a - b}; a, b > 0 \end{array}$

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The circle passing through the intersection of the circles, $x^{2} + y^{2} - 6 x = 0$ and $x^{2} + y^{2} - 4 y$ $= 0$ , having its centre on the line, $2 x - 3 y + 12 = 0$ , also passes through the point:

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The circle passing through the intersection of the circles, x2+y2-6x=0 and x2+y2-4y =0 , having its centre on the line, 2x-3y+12=0 , also passes through the point:

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The circle passing through the intersection of the circles, $x^{2} + y^{2} - 6 x = 0$ and $x^{2} + y^{2} - 4 y$ $= 0$ , having its centre on the line, $2 x - 3 y + 12 = 0$ , also passes through the point: