Circle Ques 68
- If the tangent at $(1,7)$ to the curve $x^{2}=y-6$ touches the circle $x^{2}+y^{2}+16 x+12 y+c=0$, then the value of $c$ is
(a) 195
(b) 185
(c) 85
(d) 95
(2018 Main)
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Answer:
Correct Answer: 68.(d)
Solution:
Formula:
Key Idea Equation of tangent to the curve
$x^{2}=4 a y$ at $\left(x _1, y _1\right)$ is $x x _1=4 a \quad (\frac{y+y _1}{2})$
Tangent to the curve $x^{2}=y-6$ at $(1,7)$ is
$ x=\frac{y+7}{2}-6 $
$\Rightarrow \quad 2 x-y+5=0 ……(i)$
Equation of circle is $x^{2}+y^{2}+16 x+12 y+c=0$
Centre $(-8,-6)$
$ r=\sqrt{8^{2}+6^{2}-c}=\sqrt{100-c} $
Since, line $2 x-y+5=0$ also touches the circle.
$ \begin{array}{lc} \therefore & \sqrt{100-c}=\frac{2(-8)-(-6)+5}{\sqrt{2^{2}+1^{2}}} \\ \Rightarrow & \sqrt{100-c}=\frac{-16+6+5}{\sqrt{5}} \\ \Rightarrow & \sqrt{100-c}=|-\sqrt{5}| \\ \Rightarrow & 100-c=5 \\ \Rightarrow & c=95 \end{array} $
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