Testing a Point

Video lesson1 worked example

Theory

Given a circle with centre $(a,b)$ and radius $r$ units, we can determine whether a point $(p,q)$ lies within the circle, on the circumference of the circle or out with the circle.

(p-a)^2+(q-b)^2 < r^2

Point lies within the circle

(p-a)^2+(q-b)^2 = r^2

Point lies on the circumference of the circle

(p-a)^2+(q-b)^2 > r^2

Point lies out with the circle

⚠️ Common Examiner Traps

Compare distance with radius: work out the distance from the centre to the point, then compare. Less than the radius is inside, equal is on, greater is outside.
State the conclusion: a calculation with no sentence does not answer a question that asks whether the point lies on the circle.
Substituting works too: put the coordinates into the left-hand side and compare with $r^2$ — often quicker than a square root.
Compare like with like: if you substitute, you are comparing with $r^2$ , not with $r$ .

Worked examples

Example 1

A circle has equation $(x-2)^2+(y+5)^2=29$ . Determine where the points $(2,1)$ , $(7,-3)$ and $(2,-4)$ lie in relation to the circle.

Evaluate $(x-2)^2+(y+5)^2$ for each point and compare with 29.

For (2, 1):

(2-2)^2 + (1+5)^2 = 0^2 + 6^2 = 36

Since $36 > 29$ , the point (2, 1) lies outside the circle.

For (7, -3):

(7-2)^2 + (-3+5)^2 = 5^2 + 2^2 = 25 + 4 = 29

Since $29 = 29$ , the point (7, -3) lies on the circumference of the circle.

For (2, -4):

(2-2)^2 + (-4+5)^2 = 0^2 + 1^2 = 1

Since $1 < 29$ , the point (2, -4) lies inside the circle.