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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?

1 Answer

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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