Tangent - point of contact

Question

Question: Tangent to the curve $y = x^2 + 6$ at 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:

1) $(-6, -11)$

2) $(-9, -13)$

3) $(-10, -15)$

4) $(-6, -7)$

Attempt so far: By using differentiation, I found the slope of the tangent at point $P$, and then using the point, found the equation of the tangent. However, I couldn't solve it simultaneously with the equation of the circle as $c$ is a variable. Please help!

Answer 1

HINT:

As you have already calculated the equation of the tangent

Now

Method $\#1:$

Find the ordinate/ abscissa of the intersection. For tangency, the two value must coincide.

Method $\#2:$

The perpendicular distance of the center of the circle from the tangent =raidus

Answer 2

According to the suggested, using the info from (Tangent to the curve $y = x^2 + 6$ at point P(1, 7)), slope of the tangent = … = 2

Slope of that tangent to the circle is given by $\dfrac {dy}{dx} = … = \dfrac {-(x + 8)}{y + 6}$

Since they are the same line (and thus same slope), then $\dfrac {-(x + 8)}{ y + 6} = 2$

After simplification, we have $x + 2y = –20$

There will be quite trouble to go any further because c is unknown. However, this is an MC question, the only option that fits is (-6, -7).