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

Imagine you are standing on a coordinate plane, looking at a beautiful, upward-opening parabola defined by $y=x^2+6$. You have an anchor point, $P(1,7)$, sitting right on that curve.

This is where our journey begins. We are tasked with finding a tangent line at this point, which then acts as a bridge to a mysterious circle.

Our First Step

To find the tangent, we need its slope. Think of the derivative as a slope-finding machine. When we differentiate $y=x^2+6$, we get:

By plugging in the $x$-coordinate of our point $P$, which is $1$, we find the slope $m_t=2(1)=2$.

Now, armed with the point $(1,7)$ and the slope $2$, we use the point-slope form: $y-7=2(x-1)$. Simplifying this, we get the equation of our tangent line:

This line is our path forward.

Now, let's turn our attention to the circle: $x^2+y^2+16x+12y+c=0$. To understand its position, we need its center.

By comparing this to the general form $x^2+y^2+2gx+2fy+c=0$, we identify the center $C$ at $(-g,-f)$, which is $(-8,-6)$. This center is the heart of our circle, and it holds the secret to finding the point of tangency $Q$.

Here is the crucial geometric connection: the radius $CQ$ drawn to the point of tangency $Q$ is always perpendicular to the tangent line. This makes the line $CQ$ a normal line.

Since the tangent has a slope of $2$, the normal line must have a slope $m_n$ that satisfies $m_t\times m_n=-1$. Thus, $m_n=-\frac{1}{2}$.

Now, we construct the equation of this normal line passing through $C(-8,-6)$ with slope $-\frac{1}{2}$:

Simplifying this, we get $2(y+6)=-(x+8)$, which rearranges to:

We have two lines: the tangent $2x-y+5=0$ and the normal $x+2y+20=0$. The point $Q$ is where they meet.

From the tangent equation, we know $y=2x+5$. Substituting this into the normal equation:

Expanding this, we get $x+4x+10+20=0$, or $5x+30=0$. Solving for $x$, we find $x=-6$.

Plugging this back into $y=2x+5$, we get $y=2(-6)+5=-7$.

And there it is: the coordinates of $Q$ are $(-6,-7)$. You have successfully navigated the geometry of the parabola and the circle!