Q93 / 160 · AP EAMCET 2023 (17 May Shift 2) · Mathematics

The point of intersection of the tangents drawn at the points where the line $2 x-y+3=0$ meets the circle $x^2+$ $y^2-4 x-6 y+4=0$ is

  • A
    $\left(-8, \frac{15}{2}\right)$
  • Correct answer
    B
    $\left(\frac{-5}{2}, \frac{21}{4}\right)$
  • C
    $\left(\frac{5}{2}, \frac{-21}{4}\right)$
  • D
    $\left(8, \frac{-15}{2}\right)$

Solution

Given: $x^2+y^2-4 x-6 y+4=0$ $\Rightarrow(x-2)^2+(y-3)^2=3^2$ The given equation of line is $2 x-y+3=0$ The required point is $\begin{aligned} & =\left(2-\frac{3^2 \times 2}{2 \times 2-1 \times 3+3}, 3-\frac{3^2 \times(-1)}{2 \times 2-1.3+3}\right) \\ & =\left(2-\frac{9.2}{4}, 3+\frac{9}{4}\right)=\left(-\frac{5}{2}, \frac{21}{4}\right)\end{aligned}$

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

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