Question

The equation of a circle is (x + 12)2 + (y + 16)2 = (r1)2, and the circle passes through the origin. The equation of the circle then changes to (x – 30)2 + (y – 16)2 = (r2)2, and the circle still passes through the origin. What are the values of r1 and r2?

Answer 1

For an equation to pass through the origin, (0,0) must be a solution of the equation. So, for the first one, 

$(x + 12)^{2} + (y + 16)^{2} = r_{1}^{2}$

(0, 0) must satisfy the equation and we can solve for the value of r₁ as shown below.

$(0 + 12)^{2} + (0 + 16)^{2} = r_{1}^{2}$
$12^{2} + 16^{2} = r_{1}^{2}$
$r_{1} = \sqrt{12^{2} + 16^{2}}$
$r_{1} = 20$

Since the second equation of the circle also passes through the origin, we use the same steps from the first one.

$(0 - 30)^{2} + (0 - 16)^{2} = r_{2}^{2}$
$r_{2} = \sqrt{30^{2} + 16^{2}}$
$r_{2} = 34$

Thus, we have the values for r₁ and r₂ as 20 and 34, respectively.

Answer: r₁ = 20 and r₂ = 34