Question 88·Hard·Geometry and Trigonometry

A line is tangent to a circle exactly when the distance from the circle’s center to the line equals the circle’s radius. Use the point-to-line distance formula with the center $(1,-2)$ and your general line.

After plugging into the distance formula and setting it equal to the radius $\sqrt{13}$, you will get an equation involving $m$. Square both sides carefully and simplify to get a quadratic equation in $m$.

Once you have a quadratic $am^2 + bm + c = 0$ whose roots are the slopes of the two tangents, remember that the sum of the roots equals $-b/a$. Use this to find the sum without solving for each slope separately.

In Desmos, type the quadratic equation you derived for the slopes as a function: y = 23x^2 - 96x + 51. Here, the x-values where this graph crosses the x-axis represent the two possible slopes of the tangent lines.

Zoom and pan until you can clearly see where the parabola intersects the x-axis. Tap each x-intercept to see its x-coordinate; these two x-coordinates are the two slopes $m_1$ and $m_2$.

Either add the two x-coordinates manually, or label the intercept points (for example as A and B) and enter a new expression like A.x + B.x. The result shown is the sum of the slopes of the two tangent lines.

The circle has center $(1,-2)$ and radius $\sqrt{13}$, so its equation is

You are given an external point $(7,6)$, and you are looking for the two lines through this point that just touch the circle (are tangent). Their slopes will be two different numbers; you need the sum of those slopes.

Let the slope of a line through $(7,6)$ be $m$. Then its equation in point-slope form is

Rewrite this in standard form $Ax + By + C = 0$ (needed for the distance formula):

So for this line, $A = m$, $B = -1$, and $C = 6 - 7m$.

For a line $Ax + By + C = 0$ and a point $(x_0,y_0)$, the distance from the point to the line is

Here, the point is the center of the circle $(1,-2)$, the line is $mx - y + (6 - 7m) = 0$, and the radius is $\sqrt{13}$. Plug in:

• Numerator:

• Denominator:

Set the distance equal to the radius:

Square both sides (which removes the absolute value and square root):

Start from

Multiply both sides by $m^2 + 1$:

Expand each side:

• Left side:

• Right side:

Set them equal and bring all terms to one side:

So the two possible slopes $m_1$ and $m_2$ are the two roots of the quadratic equation

For a quadratic equation $am^2 + bm + c = 0$ with roots $m_1$ and $m_2$, the sum of the roots is

Here, $a = 23$ and $b = -96$, so

Therefore, the sum of the slopes of the two tangent lines is $\dfrac{96}{23}$.