Find the center (h, k) and radius r of the circle with the given equation.?

Find the center (h, k) and radius r of the circle with the given equation.

x2 + y2 + 18x + 8y + 97 = 25

A) (h, k) = (9, 4); r = 25

B) (h, k) = (4, 9); r = 25

C) (h, k) = (-4, -9); r = 5

D) (h, k) = (-9, -4); r = 5

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  • 9 years ago
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    If a circle has centre (h,k) and radius r, then (x-h)² + (y-k)² = r². We need to find a way of converting the equation into this form.

    Have you met the concept of "completing the square"? You need to do this twice here, once for x and once for y.

    Rewriting the equation as

    x² + 18x + y² + 8y + 97 = 25.

    What value of c makes the trinomial x² + 18x + c a perfect square? In general, if you have a trinomial of the form x² + bx + c, it will be a perfect square if c = (-b/2)². So x²+18x+cwill be a perfect square trinomial if c = (-18.2)² = (-9)² = 81.

    Similarly, y² + 8y + k is a perfect square if k = (-8/2)² = (-4)² = 16. So you need to add 81 and 16 to the left hand side of the equation. But you must also subtract these numbers off so you don't change it's value!! to leave

    x² + 18x + 81 + y² + 8y + 16 + 97 - 81 - 16 = 25.

    By a stroke of luck (or more likely, clever construction of the question), 97-81-16 = 0, reducing the equation to

    x² + 18x + 81 + y² + 8y + 16 = 25.

    Noting that x²+18x+81 = (x+9)² and y²+8y+16 = (y+4)², we can write this as

    (x+9)² + (y+4)² = 25.

    Writing 25 = 5², we have

    (x+9)² + (y+4)² = 5², which is in the form (x-h)² + (y-k)² = r² with h=-9, k=-4, r = 5.

    So (h, k) = (-9, -4); r = 5.

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