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