Question

circle j has a center of (10, -3) and a radius of 9 units. circle k has a center of (2, 12). circle j and circle k intersect at exactly one point. which expression can be used to determine the radius, in units, of circle k?

a. \\(\sqrt{(12 - 9)^2 + (2 - 9)^2}\\)

b. \\(\sqrt{(-3 - 12)^2 + (10 - 2)^2}\\)

c. \\(\sqrt{(-3 - 12)^2 + (10 - 2)^2} - 9\\)

d. \\(\sqrt{(-3 - 12)^2 + (10 - 2)^2} - \sqrt{(12 - 9)^2 + (2 - 9)^2}\\)

Explanation:

Step1: Find distance between centers

The distance \(d\) between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(d = \sqrt{(y_1 - y_2)^2+(x_1 - x_2)^2}\). For centers \((10, - 3)\) (Circle J) and \((2,12)\) (Circle K), \(d=\sqrt{(-3 - 12)^2+(10 - 2)^2}\).

Step2: Analyze intersection condition

Since circles intersect at one point (tangent), \(|r_J\pm r_K|=d\). Given \(r_J = 9\), we solve for \(r_K\). If externally tangent, \(r_J + r_K=d\Rightarrow r_K=d - r_J\); if internally tangent, \(|r_K - r_J|=d\), but here \(d=\sqrt{(-3 - 12)^2+(10 - 2)^2}\), and the expression for \(r_K\) is \(d-9=\sqrt{(-3 - 12)^2+(10 - 2)^2}-9\), which matches option C.

Answer:

C. \(\sqrt{(-3 - 12)^{2}+(10 - 2)^{2}}-9\)