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BITSAT 2016 Paper

Section: Mathematics

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Question : 115 of 150

Marks: +1, -0

The lengths of the tangent drawn from any point on the circle

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

to the two circles

5x^{2}+5y^{2}-24x+32y+75=0

5x^{2}+5y^{2}-48x+64y+300=0

are in the ratio of

$1:2$
$2:3$
$3:4$
None

Solution:

Let P(h, k) be a point on the circle

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

Then the lengths of the tangents from P(h, k) to

5x^{2}+5y^{2}-24x+32y+75=0

and

5x^{2}+5y^{2}-48x+64y+300=0

are

PT_{1}=\sqrt{h^{2}+k^{2}-\frac{24}{5}h+\frac{32}{5}k+15}

and

PT_{2}=\sqrt{h^{2}+k^{2}-\frac{48}{5}h+\frac{64}{5}k+60}

or

PT_{1}=\sqrt{\frac{48}{15}h-\frac{64}{15}k-\frac{24}{5}h+\frac{32}{5}k+15}=\sqrt{\frac{32}{15}k-\frac{24}{15}h+15}

(Since (h, k) lies on

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

\therefore h^{2}+k^{2}-\frac{48}{15}h+\frac{64}{15}k=0

and

PT_{2}=\sqrt{\frac{48}{15}h-\frac{64}{15}k-\frac{48}{5}h+\frac{64}{5}k+60}

=\sqrt{-\frac{96}{15}h+\frac{128}{15}k+60}=2\sqrt{-\frac{24}{15}h+\frac{32}{15}k+15}=2PT_{1}

\Rightarrow PT_1:PT_{23}=1:2

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