Question bank / Circles

Math · Geometry and TrigonometryHard

The circle with equation x2+y2−6x+8y=0x^2 + y^2 - 6x + 8y = 0 has center (h,k)(h, k) and radius rr. What is the value of h+k+rh + k + r?

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Answer: 4

Complete the square: (x2−6x+9)+(y2+8y+16)=0+9+16(x^2 - 6x + 9) + (y^2 + 8y + 16) = 0 + 9 + 16, so (x−3)2+(y+4)2=25(x - 3)^2 + (y + 4)^2 = 25. The center is (3,−4)(3, -4) and the radius is 55. So h+k+r=3+(−4)+5=4h + k + r = 3 + (-4) + 5 = 4.

Answering 66 uses the center (−3,4)(-3, 4), but (x−3)2(x - 3)^2 means h=3h = 3 and (y+4)2(y + 4)^2 means k=−4k = -4.

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