Mathematics · Ch 6 — Circle

Diameter Form

6.1.3

Diameter Form

Setting up the picture. Let $A(x_1,y_1)$ and $B(x_2,y_2)$ be the two endpoints of a diameter of a circle, so the centre $C$ is the midpoint of $AB$ . Let $P(x,y)$ be any other point on the circle. Since $AB$ is a diameter, the angle $\angle APB$ that it subtends at any point $P$ on the circle is an angle inscribed in a semicircle — and an angle inscribed in a semicircle is always $90^\circ$ . So

AP \perp BP

Turning perpendicularity into an equation. The slope of $AP$ is $\dfrac{y-y_1}{x-x_1}$ and the slope of $BP$ is $\dfrac{y-y_2}{x-x_2}$ . Since $AP\perp BP$ , the product of these two slopes must be $-1$ :

\frac{y-y_1}{x-x_1}\times\frac{y-y_2}{x-x_2} = -1

Cross-multiplying,

(y-y_1)(y-y_2) = -(x-x_1)(x-x_2)

i.e.

(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0

This is the diameter form of the equation of a circle. It is especially convenient whenever a problem directly hands you the two endpoints of a diameter (or lets you find them), since no separate step to locate the centre or radius is needed.

Solved Example 1 — circle with centre at the origin, radius 3

This is a direct application of the standard form $x^2+y^2=r^2$ with $r=3$ : …

Figure Fig.6.3Fig. 6.3 — diameter AB with a point P on the circle

What this figure shows. A circle with A(x1, y1) and B(x2, y2) marked as the two endpoints of a diameter, centre C between them, and a third point P(x, y) on the circle joined to both A and B — used to show angle APB is 90 degrees and hence that the slopes of AP and BP multiply to -1. …

Misc Ex.1Circle with centre at origin and radius 3

Worked out. A direct one-line substitution into the standard form x^2+y^2=r^2 with r=3. Working through this worked example after reading the theory above helps consolidate the method before attempting the exercise questions that follow it in the textbook. …

Misc Ex.2Circle with centre (-3, 1) through the point (5, 2)

Worked out. Finds r^2 as the squared distance from the given centre to the given point via the distance formula, then substitutes into the centre-radius form and expands to the general form. …

Misc Ex.3Circle with A(2, -3) and B(-3, 5) as diameter endpoints

Worked out. A direct substitution into the diameter form, followed by expanding the product of the two linear factors in x and in y. Working through this worked example after reading the theory above helps consolidate the method before attempting the exercise questions that follow it in the textbook. …

Figure Fig.6.4Fig. 6.4 — circle touching the Y-axis at (0, 3)

What this figure shows. A circle with centre C(-3, 3) drawn tangent to the Y-axis at the point (0, 3), showing that the radius equals the horizontal distance from the centre to the axis of tangency. …

Misc Ex.4Circle touching the Y-axis at (0, 3) with centre (-3, 3)

Worked out. Reads the radius directly off the picture (distance from the centre to the point of tangency = 3) and substitutes into the centre-radius form. …

Figure Fig.6.5Fig. 6.5 — perpendicular distance from the centre to a chord

What this figure shows. A circle with centre C(3, -4); the line 3x-4y-5=0 cuts the circle at A and B with AB=6; the foot of the perpendicular from C to line AB is M, and the right triangle AMC (with AM=3, half of AB) is shown, used to find CA = radius via Pythagoras once CM is computed. …

Misc Ex.5Circle with centre (3, -4) where a line cuts a chord of length 6

Worked out. Computes the perpendicular distance CM from the centre to the given line, then uses the right triangle formed with half the chord length AM=3 to get the radius CA via the Pythagorean theorem, and finally substitutes into the centre-radius form. …

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