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BITSAT 2016
Exam
BITSAT
Year
2016
Subject
Mathematics
Chapter
Circle

BITSAT 2016 · Mathematics · Circle

The lengths of the tangent drawn from any point on the circle 15 x^...

The lengths of the tangent drawn from any point on the circle 15x2+15y2−48x+64y=015 x^{2}+15 y^{2}-48 x+64 y=0 to the two circles 5x2+5y2−24x+32y+75=05 x^{2}+5 y^{2}-24 x+32 y+75=0 and 5x2+5y2−48x+64y+300=05 x^{2}+5 y^{2}-48 x+64 y+300=0 are in the ratio of

Options

  1. A1: 2
  2. B2: 3
  3. C3: 4
  4. DNone

Correct answer

A. 1: 2

Step-by-step solution

Let (P(h,k)beapointonthecircle\mathrm{P}(\mathrm{h}, \mathrm{k}) be a point on the circle15 x^{2}+15 y^{2}-48 x+64 y=0Thenthelengthsofthetangentsfrom(Then the lengths of the tangents from (\mathrm{P}(\mathrm{h}, \mathrm{k}) to (5x2+5y2−24x+32y+75=05x2+5y2−48x+64y+300=0 are PT1=h2+k2−245h+325k+15 and PT2=h2+k2−485h+645k+60 or PT1=4815h−6415k−245h+325k+15=3215k−2415h+15\begin{aligned}&5 x^{2}+5 y^{2}-24 x+32 y+75=0 \\ &5 x^{2}+5 y^{2}-48 x+64 y+300=0 \text{ are }\\ &P T_{1}=\sqrt{h^{2}+k^{2}-\frac{24}{5} h+\frac{32}{5} k+15}\\ &\text{ and }P T_{2}=\sqrt{h^{2}+k^{2}-\frac{48}{5} h+\frac{64}{5} k+60}\\ &\text{ or }P T_{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} \end{aligned}(Since(((Since ((\mathrm{h}, \mathrm{k})lieson(15\mathrmx2−15\mathrmy2−48lies on (15 \mathrmx^{2}-15 \mathrmy^{2}-48\mathrm{x}+6464\mathrm{y}=0)0)(∴\mathrmh2+\mathrmk2−4815 h+6415k=0) and  PT2=4815 h−645k−485 h+645k+60=−9615 h+12815k+60=2−2415 h+3215k+15=2PT1⇒PT1:PT2=1:2\begin{aligned}&\left.\therefore \mathrmh^{2}+\mathrmk^{2}-\frac{48}{15} \mathrm{~h}+\frac{64}{15} \mathrm{k}=0\right) \\ &\text{\text{ and } } \mathrm{PT}_{2}=\sqrt{\frac{48}{15} \mathrm{~h}-\frac{64}{5} \mathrm{k}-\frac{48}{5} \mathrm{~h}+\frac{64}{5} \mathrm{k}+60}\\ &=\sqrt{-\frac{96}{15} \mathrm{~h}+\frac{128}{15} \mathrm{k}+60}=2 \sqrt{-\frac{24}{15} \mathrm{~h}+\frac{32}{15} \mathrm{k}+15}=2 \mathrm{PT}_{1}\\ &\Rightarrow \mathrm{PT}_{1}: \mathrm{PT}_{2}=1: 2 \end{aligned})
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