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aksik [14]
2 years ago
5

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?
Mathematics
1 answer:
Furkat [3]2 years ago
7 0
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
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