Circles | Sec 3 & 4 A Maths Study Guide

Secondary 3 & 4 Additional Mathematics · Study guide

Use the distance definition of a circle to connect its equation, centre, radius, intersections and tangents.

Concept → worked examples → practice20 min read · practice linked below

A circle is a fixed distance from its centre

This shared Secondary 3 and 4 guide builds circle equations from distance, then uses straight-line geometry to find intersections, tangents and unknown centres. For a circle with centre $C(a,b)$ and radius $r>0$ , every point $(x,y)$ satisfies

(x-a)^2+(y-b)^2=r^2.

This is the distance formula squared. The right side is the square of the radius. Expanding produces $x^2+y^2+2gx+2fy+c=0$ , with centre $(-g,-f)$ and $r^2=g^2+f^2-c$ . A genuine circle needs $r^2>0$ : zero gives a single point and a negative value gives no real points.

Example 1: recover the centre and radius

Find the centre and radius of $x^2+y^2-6x+4y-12=0$ , and classify $P(4,0)$ .

Complete the square separately in x and y:

(x-3)^2-9+(y+2)^2-4-12=0,\qquad (x-3)^2+(y+2)^2=25.

The centre is $(3,-2)$ and the radius is $5$ . At $P$ , the squared distance from the centre is $(4-3)^2+(0+2)^2=5<25$ , so P is inside. Equality would mean on the circumference; a value greater than 25 would mean outside.

The extreme coordinates are $x=3\pm5$ and $y=-2\pm5$ . Thus the rightmost point is $(8,-2)$ and the highest point is $(3,3)$ .

Build the equation from geometric information

If the endpoints of a diameter are known, its midpoint is the centre and half its length is the radius. If a chord is known, the centre lies on its perpendicular bisector. Two different chord bisectors locate a unique centre when the three points are noncollinear.

Example 2: a diameter determines the circle

The endpoints of a diameter are $A(-1,2)$ and $B(5,10)$ . Find the circle equation.

The midpoint is $C(2,6)$ . The diameter length is $\sqrt{6^2+8^2}=10$ , so $r=5$ . Hence

(x-2)^2+(y-6)^2=25,\qquad x^2+y^2-4x-12y+15=0.

At either endpoint, the squared distance from $C$ is $3^2+4^2=25$ , confirming both lie on the circle.

Example 3: a circle through three points

Find the circle through $A(0,0)$ , $B(4,0)$ and $D(0,6)$ .

Chord $AB$ is horizontal, with midpoint $(2,0)$ ; its perpendicular bisector is $x=2$ . Chord $AD$ is vertical, with midpoint $(0,3)$ ; its perpendicular bisector is $y=3$ . Therefore the centre is $(2,3)$ .

Using A, $r^2=2^2+3^2=13$ . The equation is $(x-2)^2+(y-3)^2=13$ , or $x^2+y^2-4x-6y=0$ . Substitution of B and D also gives zero. The three points are not collinear, so these two bisectors determine a circle.

Intersect a circle and a line

At an intersection, both equations hold. Substitute the line equation into the circle equation to obtain a quadratic. Two distinct real roots give two intersections, a repeated root gives tangency, and no real roots give no intersection. For a vertical line $x=k$ , substitute k directly and solve for y.

Example 4: find both intersection coordinates

Find the intersections of $(x-2)^2+(y-1)^2=25$ and $y=x-2$ .

(x-2)^2+(x-3)^2=25

2x^2-10x-12=0,\qquad (x-6)(x+1)=0.

Thus $x=6$ or $-1$ . Substituting in the line gives $(6,4)$ and $(-1,-3)$ . Check the circle: the squared displacements are $4^2+3^2=25$ and $(-3)^2+(-4)^2=25$ .

Explore: a secant becomes a tangent

For $x^2+y^2=25$ and $y=k$ , the intersection equation is $x^2=25-k^2$ . Predict what changes at $k=5$ and $k=-5$ .

Height k:3

Use the slider or arrow keys to move the horizontal line from k = −6 to k = 6.

The distance from the centre to y = k is |k|. Compare it with radius 5.

k = 3. Two intersections: (−4, 3) and (4, 3). The line is a secant.

Tangents are perpendicular to radii

A tangent at P is perpendicular to the radius CP. First confirm that P lies on the circle. Then find the radius gradient and use its negative reciprocal. If the radius is horizontal, the tangent is vertical; if the radius is vertical, the tangent is horizontal.

This perpendicular line meets the circle only at P: for any other point Q on that line, Pythagoras gives $CQ^2=CP^2+PQ^2>r^2$ , so Q is outside. That explains why perpendicularity identifies the tangent.

Example 5: a tangent at a known point

Find the tangent to $(x-2)^2+(y-1)^2=25$ at $P(6,4)$ .

P lies on the circle because $4^2+3^2=25$ . The centre is $C(2,1)$ , so $m_{CP}=3/4$ and the tangent gradient is $-4/3$ . Hence

y-4=-\frac43(x-6),\qquad 4x+3y=36.

The tangent passes through P. Its direction $(3,-4)$ is perpendicular to the radius direction $(4,3)$ , since $3(4)+(-4)(3)=0$ . At the rightmost point $(7,1)$ , the tangent would instead be the vertical line $x=7$ .

The tangent at P is perpendicular to CP and passes through P. The right-angle marker shows the condition that identifies this tangent.

Example 6: recover a circle from its centre and tangent

A circle has centre $C(1,2)$ and tangent $3x+4y=36$ . Find its point of contact and equation.

The tangent gradient is $-3/4$ , so the radius through the contact point has gradient $4/3$ . Its equation is $y-2=\frac43(x-1)$ , or $4x-3y= -2$ .

Solve this with $3x+4y=36$ : multiply the first by 4 and the second by 3, then add to get $25x=100$ . Thus $x=4$ and $y=6$ . The contact point is $T(4,6)$ .

Now $r^2=CT^2=3^2+4^2=25$ . The circle is $(x-1)^2+(y-2)^2=25$ . The perpendicular radius meets the given line at T and has length 5, confirming tangency.

If a circle's centre lies on a given line and its tangent at a known point is given, intersect that centre-line with the perpendicular through the contact point. Once the centre is found, the distance to the contact point supplies the radius. Always distinguish a tangent's gradient from the radius gradient.

Common mistakes and useful checks

Reading the centre with the wrong signs. $(y+2)^2$ corresponds to centre y-coordinate $-2$ .
Calling 25 the radius. In centre-radius form it is $r^2$ , so the radius is 5.
Giving only x-coordinates. Substitute every root into the line and report the coordinate pairs.
Using the original circle equation's coefficients in a discriminant. Form the intersection quadratic first.
Finding a tangent at a point outside the circumference as if it were a contact point. Substitute the proposed contact point before using the perpendicular radius.
Rounding the radius too early. Keep $r^2$ exact when writing the circle equation.

Practise coordinate geometry of circles

Start with the Secondary 3 circles questions, then use the Secondary 4 circles questions for revision. Review straight-line coordinate geometry if perpendicular bisectors or simultaneous equations need more practice.