Q91 / 160 · TS EAMCET 2013 · Mathematics

If the length of the tangent from $(h, k)$ to the circle $x^2+y^2=16$ is twice the length of the tangent from the same point to the circle $x^2+y^2+2 x+2 y=0$, then

  • A
    $h^2+k^2+4 h+4 k+16=0$
  • B
    $h^2+k^2+3 h+3 k=0$
  • Correct answer
    C
    $3 h^2+3 k^2+8 h+8 k+16=0$
  • D
    $3 h^2+3 k^2+4 h+4 k+16=0$

Solution

Given equations of circle is $$ x^2+y^2=16 $$ and $\quad x^2+y^2+2 x+2 y=0$ According to the questions, Length of the tangent from $(h, k)$ to the circle $$ x^2+y^2=16 $$ $=2 \times$ Length of the tangent from $(h, k)$ to the circle $x^2+y^2+2 x+2 y=0$ $$ \Rightarrow \sqrt{h^2+k^2-16}=2 \times \sqrt{h^2+k^2+2 h+2 k} $$ On squaring both sides, we get $$ \begin{aligned} & h^2+k^2-16=4 \times\left(h^2+k^2+2 h+2 k\right) \\ \Rightarrow \quad & 3 h^2+3 k^2+8 h+8 k+16=0 \end{aligned} $$

Hence, the correct answer is option: 'C'.

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