Question

If the circles \[ x^2+y^2-2x-8y+17=r \quad \text{and} \quad x^2+y^2-26x-18y+234=0 \] intersect at exactly one point, then the sum of all possible values of \(r\) is _______

Hint:

When two circles intersect at exactly one point, always check both internal and external tangency cases.

Answer 1

Correct Answer: 370

Concept:
A circle \(x^2+y^2+2gx+2fy+c=0\) has center \((-g,-f)\) and radius \(\sqrt{g^2+f^2-c}\).
Two circles intersect at exactly one point if they touch each other (are tangent).
For tangency: \[ d = r_1 + r_2 \quad \text{or} \quad d = |r_1 - r_2|, \] where \(d\) is the distance between the centers.
Step 1: Find the center and radius of the first circle \[ x^2+y^2-2x-8y+(17-r)=0 \] Center: \[ C_1(1,4) \] Radius: \[ r_1=\sqrt{1^2+4^2-(17-r)}=\sqrt{r} \]
Step 2: Find the center and radius of the second circle \[ x^2+y^2-26x-18y+234=0 \] Center: \[ C_2(13,9) \] Radius: \[ r_2=\sqrt{13^2+9^2-234}=\sqrt{16}=4 \]
Step 3: Find the distance between the centers \[ d=\sqrt{(13-1)^2+(9-4)^2}=\sqrt{144+25}=13 \]
Step 4: Apply the tangency conditions (i) External tangency: \[ 13=\sqrt{r}+4 \Rightarrow \sqrt{r}=9 \Rightarrow r=81 \] (ii) Internal tangency: \[ 13=|\sqrt{r}-4| \Rightarrow \sqrt{r}=17 \Rightarrow r=289 \]
Step 5: Sum of all possible values of \(r\) \[ 81+289=370 \]
Final Answer: \(\boxed{370}\)