Key Concepts and Formulas

• Equation of Tangent to a Curve: Implicit differentiation can be used to find the slope of the tangent. For a curve $f(x, y) = 0$, the slope $dy/dx$ can be found and then the tangent line equation is $y - y_1 = m(x - x_1)$.
• General Equation of a Circle: The general equation of a circle is $x^2 + y^2 + 2gx + 2fy + c = 0$. Its center is $(-g, -f)$, and its radius is $r = \sqrt{g^2 + f^2 - c}$.
• Condition for Tangency of a Line to a Circle: A line $Ax + By + C = 0$ is tangent to a circle if and only if the perpendicular distance from the center of the circle to the line is equal to the radius of the circle. The distance $d$ from a point $(x_0, y_0)$ to a line $Ax + By + C = 0$ is given by the formula: $d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$.

Step-by-Step Solution

Step 1: Finding the Equation of the Tangent Line to the Curve

The given curve is $x^2 = y - 6$. We need to find the tangent line at the point $(1, 7)$.

• Why this step? We need the equation of the line that touches the circle. This line is the tangent to the given curve at the specified point.

We can rewrite the equation as $y = x^2 + 6$. Differentiating with respect to $x$, we get: $\frac{dy}{dx} = 2x$ At the point $(1, 7)$, the slope of the tangent is: $m = \frac{dy}{dx}\Big|_{x=1} = 2(1) = 2$ The equation of the tangent line is given by: $y - y_1 = m(x - x_1)$ $y - 7 = 2(x - 1)$ $y - 7 = 2x - 2$ $y = 2x + 5$ Rewriting in the standard form $Ax + By + C = 0$: $2x - y + 5 = 0$

Step 2: Finding the Center and Radius of the Circle

The given circle equation is $x^2 + y^2 + 16x + 12y + c = 0$.

• Why this step? To apply the tangency condition, we need to know the circle's center and its radius.

Comparing this to the general equation of a circle $x^2 + y^2 + 2gx + 2fy + c = 0$, we get: $2g = 16 \implies g = 8$ $2f = 12 \implies f = 6$ The center of the circle is $(-g, -f) = (-8, -6)$. The radius of the circle is $r = \sqrt{g^2 + f^2 - c}$: $r = \sqrt{8^2 + 6^2 - c} = \sqrt{64 + 36 - c} = \sqrt{100 - c}$

Step 3: Applying the Condition for Tangency

The tangent line $2x - y + 5 = 0$ touches the circle with center $(-8, -6)$ and radius $r = \sqrt{100 - c}$.

• Why this step? This is the crucial link between the tangent line and the circle. The geometric condition of tangency is translated into an algebraic equation.

The perpendicular distance from the center of the circle to the tangent line must be equal to the radius of the circle. Let the center be $(x_0, y_0) = (-8, -6)$ and the line be $Ax + By + C = 0$, where $A=2$, $B=-1$, and $C=5$. The distance $d$ is: $d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} = \frac{|2(-8) + (-1)(-6) + 5|}{\sqrt{2^2 + (-1)^2}} = \frac{|-16 + 6 + 5|}{\sqrt{4 + 1}} = \frac{|-5|}{\sqrt{5}} = \frac{5}{\sqrt{5}} = \sqrt{5}$ Equating the distance $d$ to the radius $r$: $\sqrt{5} = \sqrt{100 - c}$ Squaring both sides: $5 = 100 - c$ Solving for $c$: $c = 100 - 5 = 95$

Common Mistakes & Tips

• Be careful with the signs when finding the center of the circle. Remember the center is $(-g, -f)$, not $(g, f)$.
• Double-check the arithmetic when calculating the distance from a point to a line.
• Remember to square both sides of the equation when eliminating the square root.

Summary

We found the equation of the tangent line to the curve, then determined the center and radius of the circle. Finally, we used the condition for tangency to relate the distance from the center to the line to the radius, allowing us to solve for the unknown constant $c$. The value of $c$ is 95.

The final answer is $\boxed{95}$, which corresponds to option (A).