Math · Precalculus · Worked example

Find the center and radius of a circle

Find the center and radius of the circle x² + y² − 6x + 4y − 12 = 0.

x^{2} + y^{2} - 6 x + 4 y - 12 = 0

Group the terms

Move the constant to the right and put the x-terms and the y-terms together.

\begin{array}{l} x^{2} + y^{2} - 6 x + 4 y - 12 = 0 \\ x^{2} - 6 x + y^{2} + 4 y = 12 \end{array}

Complete both squares

Half of −6 is −3, and (−3)² = 9; half of 4 is 2, and 2² = 4. Add 9 and 4 to both sides.

\begin{array}{l} x^{2} - 6 x + y^{2} + 4 y = 12 \\ (x^{2} - 6 x + 9) + (y^{2} + 4 y + 4) = 12 + 9 + 4 \end{array}

\begin{aligned} x^{2} - 6 x + y^{2} + 4 y = 12 \\ (x^{2} - 6 x + 9) \\ + (y^{2} + 4 y + 4) = 12 + 9 + 4 \end{aligned}

Write the squares

Each group is now a perfect square, and the right side is 25.

{(x - 3)}^{2} + {(y + 2)}^{2} = 25

Read the center and radius

Compare with (x − h)² + (y − k)² = r²: h = 3, k = −2 and r² = 25, so r = 5.

Result

Center (3, −2), radius 5.

Your turn

Find the center and radius of x² + y² + 8x − 2y + 8 = 0.

Show the answer and explanation

Center (−4, 1), radius 3.

Add 16 and 1 to both sides: x² + 8x + 16 + y² − 2y + 1 = −8 + 16 + 1, so (x + 4)² + (y − 1)² = 9.

\begin{array}{l} x^{2} + y^{2} + 8 x - 2 y + 8 = 0 \\ {(x + 4)}^{2} + {(y - 1)}^{2} = 9 \end{array}

Keep exploring

In Implicit curves & inequality regions, change = 0 to < 0: the inside of the circle is shaded, and the boundary is dashed because the circle itself is excluded.

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