Question:
If the tangent at $(1, 7)$ to the curve $x^2 = y - 6$ touches the circle $x^2 + y^2 + 16x + 12y + c = 0$ then the value of c is:
If the tangent at $(1, 7)$ to the curve $x^2 = y - 6$ touches the circle $x^2 + y^2 + 16x + 12y + c = 0$ then the value of c is:
Updated On: Jul 15, 2024
- 195
- 185
- 85
- 95
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The Correct Option is D
Solution and Explanation
Equation of tangent at $(1,7)$ to curve $x^{2}=y-6$ is
$x-1=\frac{1}{2}(y+7)-6$
$2 x-y+5=0\,\,\,\,\,\,\,...(i)$
Centre of circle $=(-8,-6)$
Radius of circle $=\sqrt{64+36-c}=\sqrt{100-c}$
$\because$ Line (i) touches the circle
$\therefore \left|\frac{2(-8)-(-6)+5}{\sqrt{4+1}}\right|=\sqrt{100-c} $
$ \sqrt{5}=\sqrt{100-c} $
$\Rightarrow c =95 $
$x-1=\frac{1}{2}(y+7)-6$
$2 x-y+5=0\,\,\,\,\,\,\,...(i)$
Centre of circle $=(-8,-6)$
Radius of circle $=\sqrt{64+36-c}=\sqrt{100-c}$
$\because$ Line (i) touches the circle
$\therefore \left|\frac{2(-8)-(-6)+5}{\sqrt{4+1}}\right|=\sqrt{100-c} $
$ \sqrt{5}=\sqrt{100-c} $
$\Rightarrow c =95 $
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