Question

Tangent to the curve $y=x^2+6$ at a point P $(1,7)$ touches the circle $x^2+y^2+16x+12y+c=0$ at a point Q. Then the coordinates of Q are

• A. $(-6,-11)$
• B. $(-9,-13)$
• C. $(-10,-15)$
• D. $(-6,-7)$

Answer 1

Answer: (D)

$(-6,-7)$

Curve, $y=x^2+6$

$=>x^2=y-6...................\left(1\right)$

Vertex:$(0,6),$ $a=\frac{1}{4}$

$T$ at $P(1,7)=>x=\frac{y+7}{2}-6$

$2x=y-5................\left(2\right)$ $(m=2)$ $(c^{\prime }=5)$

Circle: $x^2+y^2+16x+12y+c=0...................\left(3\right)$ $($radius$={\sqrt{8}}^2+6^2-c)$

If $\left(2\right)$ is tangent to $\left(3\right),$

$=>\left(radius\right)(\sqrt{1}+m^2)=c^{\prime }$

$=>(5{)}^2=[1+\left(2{)}^2\right][100-c]$

$=>5=100-c$

$=>c=95$ $=>a=\sqrt{5}$

$PO$ contact:$(\pm \frac{am}{\sqrt{1+m^2}},\pm \frac{a}{\sqrt{1+m^2}})$

:$(-6,-7).$