Q60 / 160 · AP EAMCET 2010 · Mathematics

If the lengths of tangents drawn to the circles $x^2+y^2-8 x+40=0$ $5 x^2+5 y^2-25 x+80=0$ $x^2+y^2-8 x+16 y+160=0$ From the point $P$ are equal, then $P$ is equal to

  • A
    $\left(8, \frac{15}{2}\right)$
  • B
    $\left(-8, \frac{15}{2}\right)$
  • Correct answer
    C
    $\left(8, \frac{-15}{2}\right)$
  • D
    $\left(-8, \frac{-15}{2}\right)$

Solution

Let $P\left(x_1, y_1\right)$ be the point from which the tangents are drawn to the circles $S_1 \equiv x^2+y^2-8 x+40=0$ $S_2 \equiv 5 x^2+5 y^2-25 x+80=0$ $S_3 \equiv x^2+y^2-8 x+16 y+160=0$ Since, the length of the tangent from $P$ to the circle $S_1, S_2, S_3$ are equal $\therefore \quad \sqrt{S_1}=\sqrt{S_2}=\sqrt{S_3}$ $\Rightarrow \quad S_1=S_2=S_3$ $x_1^2+y_1^2-8 x_1+40=5 x_1^2+5 x_2^2-25 x_1+80$ $=x_1^2+y^2-8 x_1+16 y_1+160$ $\ldots$ (i) Taking first and third part of above relation (i), we get $-40+16 y_1+160=0$ $16 y_1+120=0$ $y_1=-\frac{120}{16} \Rightarrow-\frac{15}{2}$ Taking first and second part of the relation (i). $-3 x_1+24=0$ $x_1=\frac{24}{3} \Rightarrow x_1=8$ Hence, the point $P$ is $\left(8, \frac{-15}{2}\right)$.

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

Relevant Questions