The lengths of the tangent drawn from any point on the circle $15 x^{2} + 15 y^{2} - 48 x + 64 y = 0$ to the two circles $5 x^{2} + 5 y^{2} - 24 x + 32 y + 75 = 0, 5 x^{2} + 5 y^{2} - 48 x + 64 y + 300 = 0$ are in the ratio of

1 : 2

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2 : 3

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3 : 4

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4 : 5

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Solution

The correct option is A $1 : 2$
Take $(0, 0)$ a point on $15 x^{2} + 15 y^{2} - 48 x + 64 y = 0$ circle

so, length of tangent from (0,0) to $5 x^{2} + 5 y^{2} - 24 x + 32 y + 75 = 0$ is

$L_{1} = \sqrt{s_{1}} = \sqrt{15}$

and length of tangent from (0,0) to $5 x^{2} + 5 y^{2} - 48 x + 64 y + 300 = 0$

$L_{2} = \sqrt{\frac{300}{5}} = \sqrt{60}$

so, $L_{1} : L_{2} = \sqrt{15} : \sqrt{60}$

$= 1 : 2$

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Similar questions

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15 x^{2} + 15 y^{2} - 48 x + 64 y = 0

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5 x^{2} + 5 y^{2} - 24 x + 32 y + 75 = 0, 5 x^{2} + 5 y^{2} - 48 x + 64 y + 300 = 0

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The lengths of the tangent drawn from any point on the circle 15x2+15y2−48x+64y=0 to the two circles 5x2+5y2−24x+32y+75=0,5x2+5y2−48x+64y+300=0 are in the ratio of

The correct option is A 1:2 Take (0,0) a point on 15x2+15y2−48x+64y=0 circleso, length of tangent from (0,0) to 5x2+5y2−24x+32y+75=0 isL1=√s1=√15and length of tangent from (0,0) to 5x2+5y2−48x+64y+300=0L2=√3005=√60so, L1:L2=√15:√60=1:2

The lengths of the tangent drawn from any point on the circle $15 x^{2} + 15 y^{2} - 48 x + 64 y = 0$ to the two circles $5 x^{2} + 5 y^{2} - 24 x + 32 y + 75 = 0, 5 x^{2} + 5 y^{2} - 48 x + 64 y + 300 = 0$ are in the ratio of