JEE Advanced · Mathematics · 12. Circle
The locus of the mid-point of the chord of contact of tangents drawn from points lying on the straight line \(4 x-\) \(5 y=20\) to the circle \(x^{2}+y^{2}=9\) is
- A \(20\left(x^{2}+y^{2}\right)-36 x+45 y=0\)
- B \(20\left(x^{2}+y^{2}\right)+36 x-45 y=0\)
- C \(36\left(x^{2}+y^{2}\right)-20 x+45 y=0\)
- D \(36\left(x^{2}+y^{2}\right)+20 x-45 y=0\)
Answer & Solution
Correct Answer
(A) \(20\left(x^{2}+y^{2}\right)-36 x+45 y=0\)
Step-by-step Solution
Detailed explanation
Any point \(P\) on line \(4 x-5 y=20\) can be considered as \(\left(\alpha, \frac{4 \alpha-20}{5}\right)\).
Equation of chord of contact \(A B\) to the circle \(x^{2}+y^{2}=9\)
drawn from point \(\mathrm{P}\left(\alpha, \frac{4 \alpha-20}{5}\right)\) is
x. \(\alpha+\) y. \(\left(\frac{4 \alpha-20}{5}\right)=9...(i)\)

Also the equation of chord \(A B\) whose mid point is \((h, k)\) is \(h x+k y=h^{2}+k^{2}....(ii)\)
\(\because \quad\) Equations (i) and (ii) represent the same line,
\(\therefore \quad \frac{h}{\alpha}=\frac{k}{\frac{4 \alpha-20}{5}}=\frac{h^{2}+k^{2}}{9}\)
\(\Rightarrow 5 k \alpha=4 h \alpha-20 h\) and \(9 h=\alpha\left(h^{2}+k^{2}\right)\)
\(\Rightarrow \alpha=\frac{20 h}{4 h-5 k} \quad\) and \(\alpha=\frac{9 h}{h^{2}+k^{2}}\)
\(\Rightarrow \quad \frac{20 h}{4 h-5 k}=\frac{9 h}{h^{2}+k^{2}} \Rightarrow 20\left(h^{2}+k^{2}\right)=9(4 h-5 k)\)
\(\therefore \quad\) Locus of \((h, k)\) is \(20\left(x^{2}+y^{2}\right)-36 x+45 y=0\)
Equation of chord of contact \(A B\) to the circle \(x^{2}+y^{2}=9\)
drawn from point \(\mathrm{P}\left(\alpha, \frac{4 \alpha-20}{5}\right)\) is
x. \(\alpha+\) y. \(\left(\frac{4 \alpha-20}{5}\right)=9...(i)\)

Also the equation of chord \(A B\) whose mid point is \((h, k)\) is \(h x+k y=h^{2}+k^{2}....(ii)\)
\(\because \quad\) Equations (i) and (ii) represent the same line,
\(\therefore \quad \frac{h}{\alpha}=\frac{k}{\frac{4 \alpha-20}{5}}=\frac{h^{2}+k^{2}}{9}\)
\(\Rightarrow 5 k \alpha=4 h \alpha-20 h\) and \(9 h=\alpha\left(h^{2}+k^{2}\right)\)
\(\Rightarrow \alpha=\frac{20 h}{4 h-5 k} \quad\) and \(\alpha=\frac{9 h}{h^{2}+k^{2}}\)
\(\Rightarrow \quad \frac{20 h}{4 h-5 k}=\frac{9 h}{h^{2}+k^{2}} \Rightarrow 20\left(h^{2}+k^{2}\right)=9(4 h-5 k)\)
\(\therefore \quad\) Locus of \((h, k)\) is \(20\left(x^{2}+y^{2}\right)-36 x+45 y=0\)
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