The lengths of the tangents from any point on the circle $15 x^{2} + 15 y^{2} - 48 x + 64 y = 0$ to the 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

Question

The lengths of the tangents from any point on  the circle $$15x2+15y2$$−$$48x+64y=0$$ to the $$circles5x2+5y2$$−$$24x+32y+75=05x2+5y2$$−$$48x+64y+300=0$$ are in the ratio

Infinity Learn · Accepted Answer

Let P (h$$, $$k) be a point on the $$circle15x2+15y2$$−$$48x+64y=0Then$$ $$15h2+15k2$$−$$48h+64k=0$$∴ $$h2+k2=4815h$$−$$6415k$$             …(i)Let$$ $$PT1$$ and $$PT2$$ be the lengths of the tangents from P $$(h$$, $$k) to $$5x2+5y2$$−$$24x+32y+75=0and$$, $$5x2+5y2$$−$$48x+64y+300=0$$ respectively. Then,$$PT1=h2+k2$$−$$245h+325k+15and$$, $$PT2=h2+k2$$−$$485h+645k+60$$⇒ $$PT1=4815h$$−$$6415k$$−$$245h+325k+15$$       [Using (i)] ⇒ $$PT1=32k15$$−$$2415h+15and$$, $$PT2=4815h$$−$$6415k$$−$$485h+645k+60$$            [Using (i)] ⇒ $$PT2=$$−$$9615h+12815k+60$$⇒ $$PT2=2$$−$$2415h+3215k+15=2PT1$$∴ $$PT1$$:$$PT2=1$$:$$2$$

The lengths of the tangents from any point on the circle $15 x^{2} + 15 y^{2} - 48 x + 64 y = 0$ to the 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

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The lengths of the tangents from any point on the circle 15x2+15y2−48x+64y=0 to the circles5x2+5y2−24x+32y+75=05x2+5y2−48x+64y+300=0 are in the ratio

Remember concepts with our Masterclasses.

Ready to Test Your Skills?

free study material

The lengths of the tangents from any point on the circle $15 x^{2} + 15 y^{2} - 48 x + 64 y = 0$ to the 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