Question

Difficulty: MediumEquations and Graphs of Circles

A circle in the standard (x,y)(x, y) coordinate plane is represented by the equation x2+y2+4x−6y−12=0x^2 + y^2 + 4x - 6y - 12 = 0. What is the distance between the two points where this circle intersects the xx-axis?

Answer: 8 units

Answer

The distance between the two points where the circle intersects the xx-axis is 8 units.
Setting y=0y = 0 gives the quadratic equation x2+4x−12=0x^2 + 4x - 12 = 0. Factoring yields (x+6)(x−2)=0(x + 6)(x - 2) = 0, giving solutions x=−6x = -6 and x=2x = 2. The distance between these two xx-intercepts is 2−(−6)=82 - (-6) = 8 units.

Step-by-Step Solution

1
Substitute y=0y = 0 into the circle equation.
x2+4x−12=0x^2 + 4x - 12 = 0
Points on the xx-axis have a yy-coordinate of 0.
2
Solve the quadratic equation for xx by factoring.
x=−6x = -6 and x=2x = 2
Factoring (x+6)(x−2)=0(x + 6)(x - 2) = 0 yields the xx-coordinates of the intersection points.
3
Find the distance between (−6,0)(-6, 0) and (2,0)(2, 0).
2−(−6)=82 - (-6) = 8
The horizontal distance between two points (x1,0)(x_1, 0) and (x2,0)(x_2, 0) is ∣x2−x1∣|x_2 - x_1|.

Key Concept

Finding xx-intercepts of a circle given in general form
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