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Difficulty: MediumEquations and Graphs of Circles

In the standard (x,y)(x, y) coordinate plane, a circle is defined by the equation x2+y2−12x+6y+20=0x^2 + y^2 - 12x + 6y + 20 = 0. What is the distance from the center of this circle to the point (9,1)(9, 1)?

Answer: 5

Answer

The distance from the center of the circle to the point (9,1)(9, 1) is 5.
Completing the square transforms x2+y2−12x+6y+20=0x^2 + y^2 - 12x + 6y + 20 = 0 into standard form (x−6)2+(y+3)2=25(x - 6)^2 + (y + 3)^2 = 25, establishing the center of the circle at (6,−3)(6, -3). Applying the coordinate distance formula between (6,−3)(6, -3) and (9,1)(9, 1) yields (9−6)2+(1−(−3))2=32+42=25=5\sqrt{(9 - 6)^2 + (1 - (-3))^2} = \sqrt{3^2 + 4^2} = \sqrt{25} = 5.

Step-by-Step Solution

1
Group the xx and yy terms and move the constant term to the right side of the equation.
(x2−12x)+(y2+6y)=−20(x^2 - 12x) + (y^2 + 6y) = -20
Grouping terms isolates the variables to prepare for completing the square.
2
Complete the square for both variable expressions.
(x−6)2+(y+3)2=−20+36+9=25(x - 6)^2 + (y + 3)^2 = -20 + 36 + 9 = 25
Adding (−12/2)2=36( -12 / 2 )^2 = 36 and (6/2)2=9( 6 / 2 )^2 = 9 to both sides converts the equation to standard form (x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2.
3
Determine the coordinates of the center (h,k)(h, k).
Center is (6,−3)(6, -3)
In the standard circle equation (x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2, the center is given by (h,k)(h, k).
4
Calculate the distance between the center (6,−3)(6, -3) and the point (9,1)(9, 1) using the distance formula.
d=(9−6)2+(1−(−3))2=32+42=25=5d = \sqrt{(9 - 6)^2 + (1 - (-3))^2} = \sqrt{3^2 + 4^2} = \sqrt{25} = 5
The distance formula d=(x2−x1)2+(y2−y1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} measures the straight-line distance between two points.

Key Concept

Equations of Circles and Coordinate Distance
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Topics

Equations and Graphs of Circles

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