A-Level Circles Practice Questions

Recall the standard equation of a circle

Every circle can be written this way, where $(a,b)$ is the centre and $r$ is the radius. Our job is just to slot the right numbers into $a$, $b$ and $r$.

Write down the centre and radius

The centre gives us $a=3$ and $b=2$, and the radius is $r=5$. Keep them separate so we do not mix up the two brackets.

Substitute into the formula

Replace $a$ with $3$ and $b$ with $2$. The right-hand side is the radius squared, so we write $5^2$ for now.

Work out the radius squared

Squaring the radius gives $5^2=25$. That finishes the equation of the circle.

Find the distance from the centre to the point

A point is on the circle when its distance from the centre equals the radius. So we measure the distance from $(2,1)$ to $(5,5)$.

Simplify inside the root

The differences are $3$ and $4$; squaring and adding gives $25$.

Compare with the radius

The distance is $5$, exactly equal to the radius, so the point is on the circle.

Group the terms

Gather $x$ and $y$ terms and move the constant to the right, ready to complete the square.

Complete the square on x

Halve $-6$ to get $-3$, write $(x-3)^2$ and subtract $9$.

Complete the square on y

Halve $4$ to get $2$, write $(y+2)^2$ and subtract $4$.

Form the standard equation

Substituting back and tidying gives $r^2=12+9+4=25$.

Find the radius

The radius is $\sqrt{25}=5$.

Double it for the diameter

The diameter is twice the radius, so $d=10$.

Understand what the question is asking

Before diving in, we read the question carefully and picture the circle. Being clear on what we must find, and what we are given, stops us wasting effort.

Gradient of the radius

The radius runs from $(0,0)$ to $(3,4)$, giving gradient $\tfrac{4}{3}$.

Gradient of the tangent

The tangent is perpendicular to the radius, so we take the negative reciprocal.

Equation of the tangent

Use the point $(3,4)$ and the tangent gradient in the straight-line formula.

Tidy the equation

Multiply through by $4$ and expand to clear the fraction.

Write in a neat form

Rearranging gives $3x+4y=25$.

Set y = 0 for the x-axis

On the $x$-axis $y=0$, so $3x=25$.

State Q

So the tangent crosses the $x$-axis at $Q\left(\tfrac{25}{3},0\right)$.

Check the answer looks sensible

It is always worth pausing to ask whether the answer is reasonable: a radius or length must be positive, and any coordinates should sit where we expect them on a sketch.

Re-read the question

A common way to lose marks is to answer a slightly different question. We re-read it and confirm we have given exactly what was requested, in the required form.

Understand what the question is asking

Before diving in, we read the question carefully and picture the circle. Being clear on what we must find, and what we are given, stops us wasting effort.

Set up the right-angled triangle OAP

The tangent $AP$ meets the radius $OA$ at $90^{\circ}$, so triangle $OAP$ is right-angled at $A$.

State the radius

The circle has $r^2=20$, so $OA=\sqrt{20}=2\sqrt5$.

Find OP

The distance from the centre $O(0,0)$ to $P(0,10)$ is simply $10$.

Find the tangent length AP (Pythagoras)

Using the right angle at $A$, $AP^2=OP^2-OA^2$.

Simplify AP

So $AP=\sqrt{80}=4\sqrt5$.

Area of one triangle OAP

Triangle $OAP$ is right-angled at $A$, so its area is half the product of the two perpendicular sides $OA$ and $AP$.

Compute the triangle area

Since $2\sqrt5\times4\sqrt5=8\times5=40$, half of that is $20$.

The kite is two such triangles

By symmetry, triangle $OBP$ is identical, and together they make the kite $OAPB$.

State the area

So the area of the kite $OAPB$ is $40$ square units.

Check the answer looks sensible

It is always worth pausing to ask whether the answer is reasonable: a radius or length must be positive, and any coordinates should sit where we expect them on a sketch.

Re-read the question

A common way to lose marks is to answer a slightly different question. We re-read it and confirm we have given exactly what was requested, in the required form.

Recall the method used

The main tool here was tangent length. Recognising which circle property a question needs is the fastest way to know how to start next time.

Link to earlier topics

Circle work leans heavily on earlier skills such as completing the square, the distance formula and perpendicular gradients. Keeping those sharp makes these questions much easier.

Watch out for sign slips

Most mistakes in circle questions come from signs: $(y+3)$ means $b=-3$, and the right-hand side is $r^2$, not $r$. A quick sign check avoids these traps.