KCSE 2025 Maths P2 Q11 — Equation of a Circle and its x-Intercepts

KCSE 2025 Form 4 Coordinate Geometry

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The Question

“A circle has centre (-2, 3) and radius 5 units, and is drawn on a Cartesian plane. Determine the x-intercepts of the circle.”

1

Recall the equation of a circle

A circle with centre (a, b) and radius r is described by the standard equation. Each point on the circle is exactly r units from the centre, which is what this equation captures using Pythagoras on the horizontal and vertical distances from the centre.

(x−a)2+(y−b)2=r2(x - a)^{2} + (y - b)^{2} = r^{2}
2

Substitute the centre and radius

Put in a equals -2, b equals 3 and r equals 5. Notice that x minus a negative two becomes x plus two, and squaring the radius five gives twenty-five. This is the equation of the given circle.

(x−(−2))2+(y−3)2=52(x - (-2))^{2} + (y - 3)^{2} = 5^{2}
(x+2)2+(y−3)2=25(x + 2)^{2} + (y - 3)^{2} = 25
3

Set y = 0 for the x-intercepts

The x-intercepts are the points where the circle crosses the x-axis, and every point on the x-axis has a y-coordinate of zero. So substitute y equals 0 into the equation. The term 0 minus 3 squared becomes 9, which you then subtract from both sides.

(x+2)2+(0−3)2=25(x + 2)^{2} + (0 - 3)^{2} = 25
(x+2)2+9=25(x + 2)^{2} + 9 = 25
(x+2)2=16(x + 2)^{2} = 16
4

Solve for x with both roots

Take the square root of both sides, remembering that a square root gives a positive and a negative value. The root of 16 is 4, so x plus 2 equals plus or minus 4. Solving each case gives the two x-values where the circle meets the axis.

x+2=±4x + 2 = \pm 4
x=−2+4=2orx=−2−4=−6x = -2 + 4 = 2 \quad \text{or} \quad x = -2 - 4 = -6

Final Result

The circle cuts the x-axis at the points (2, 0) and (-6, 0).

Why this method works

The standard circle equation is just Pythagoras in disguise: (x - a) and (y - b) are the horizontal and vertical distances from the centre, and their squares add to the square of the radius for every point on the circle. Setting y = 0 restricts attention to the x-axis, turning the two-variable circle equation into a single quadratic in x. The plus-or-minus square root is essential because a chord across the axis meets the circle at two symmetric points, one on each side of the centre's x-coordinate.

✓

Both points are 5 units from the centre (-2, 3). For (2, 0): the distance is the square root of (2 - (-2)) squared plus (0 - 3) squared, which is the root of 16 plus 9, equal to the root of 25, which is 5. The point (-6, 0) checks the same way.