Tewtor.ai

Quick question before we dive in: are you comfortable using the power-of-a-point idea for tangents to a circle? It says the tangent length from a point $A$ is $\sqrt{AP^2 - r^2}$ where $P$ is the center and $r$ the radius. If so, here’s a clean path.

Given:

• From $P(2,-2)$ tangent length is $2\sqrt{2}$, so $OP^2 - r^2 = (2\sqrt{2})^2 = 8$.
• From $Q(5,5)$ tangent length is $5\sqrt{2}$, so $OQ^2 - r^2 = (5\sqrt{2})^2 = 50$. Let the circle have center $O(h,k)$ and radius $r$.

Set up:

• $OP^2 = (h-2)^2 + (k+2)^2$
• $OQ^2 = (h-5)^2 + (k-5)^2$

So:

1. $(h-2)^2 + (k+2)^2 - r^2 = 8$
2. $(h-5)^2 + (k-5)^2 - r^2 = 50$

Subtract (1) from (2) to remove $r^2$: $(h-5)^2 - (h-2)^2 + (k-5)^2 - (k+2)^2 = 42$.

Expand differences: $[(h^2 - 10h + 25) - (h^2 - 4h + 4)] + [(k^2 - 10k + 25) - (k^2 + 4k + 4)] = 42$, $(-6h + 21) + (-14k + 21) = 42$, $-6h - 14k + 42 = 42$, $-6h - 14k = 0$, $3h + 7k = 0$.

Thus $k = -\frac{3}{7}h$.

Now use one equation to link $h,k,r$. From (1): $(h-2)^2 + (k+2)^2 - r^2 = 8$.

But we actually don’t need $r$. For the point $R(8,12)$ the tangent length squared is: $OR^2 - r^2 = [(h-8)^2 + (k-12)^2] - r^2$.

Use (1) to eliminate $r^2$: $OR^2 - r^2 = [(h-8)^2 + (k-12)^2] - \big[(h-2)^2 + (k+2)^2 - 8\big]$.

So tangent length squared from $R$ equals $T_R^2 = [(h-8)^2 - (h-2)^2] + [(k-12)^2 - (k+2)^2] + 8$.

Compute differences:

• For $h$-part: $(h^2 - 16h + 64) - (h^2 - 4h + 4) = -12h + 60$.
• For $k$-part: $(k^2 - 24k + 144) - (k^2 + 4k + 4) = -28k + 140$.

So $T_R^2 = (-12h + 60) + (-28k + 140) + 8 = -12h - 28k + 208$.

Use $3h + 7k = 0$ to remove $h$ and $k$. Multiply by $4$: $12h + 28k = 0 \Rightarrow -12h - 28k = 0$.

Therefore $T_R^2 = 0 + 208 = 208$.

Hence the tangent length from $R$ is $\sqrt{208} = 4\sqrt{13}$.

Check: clean cancellation is a good sign.

Answer: $4\sqrt{13}$.

Your turn: What was the key step that let us avoid finding the actual center and radius?