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G9 · Find a circle radius, write its equation, and sketch its graph

Learn to find a circle radius, write its equation, and sketch its graph through clear examples and targeted practice.

Ontario Grade 10 Mathematics

Analytic Geometry

Use the centre and one point on the circle to move between a diagram, an equation, and a graph.

A circle is the set of points that are the same distance from one fixed point. The fixed point is the centre. The distance from the centre to any point on the circle is the radius. In this lesson, you will use those ideas on a coordinate grid. You will find a radius, write an equation, and use the equation to guide a sketch.

What you will learn

1. Grade 9 bridge: distance on a coordinate grid

On a coordinate grid, a point is written as an ordered pair, such as (3,4)(3,4). The first number gives the horizontal position, and the second gives the vertical position. The horizontal coordinate is often called xx; the vertical coordinate is called yy.
You may already know how to find the distance between two points by using horizontal and vertical changes. For example, moving from (0,0)(0,0) to (3,4)(3,4) means moving 33 units horizontally and 44 units vertically. These changes make the legs of a right triangle. The distance between the points is the hypotenuse, the longest side of that triangle.
The Pythagorean theorem relates the lengths of the sides of a right triangle. If the horizontal change is aa and the vertical change is bb, the distance dd satisfies d2=a2+b2d^2=a^2+b^2. This is the same distance idea you will use to find a circle’s radius from its centre to a point on the circle.
d2=a2+b2d^2=a^2+b^2

2. From the centre and radius to the equation

A circle’s equation describes all the points (x,y)(x,y) that lie on it. Let the centre be (h,k)(h,k) and the radius be rr. For any point (x,y)(x,y) on the circle, the horizontal change from the centre is x−hx-h, and the vertical change is y−ky-k.
The distance from (h,k)(h,k) to (x,y)(x,y) must be the radius. Using the right-triangle distance idea gives the circle equation. The quantities inside the brackets show how far a point is from the centre in each direction. Squaring them makes the horizontal and vertical changes combine as required by the Pythagorean theorem.
Pay attention to the signs. If the centre’s horizontal coordinate is negative, subtracting it may look like addition. For example, if the centre has horizontal coordinate −2-2, the horizontal part is x−(−2)x-(-2), which is x+2x+2. The same rule applies to the vertical coordinate.
When the centre is (0,0)(0,0), the equation becomes especially simple. The circle still includes every point whose distance from the origin is the radius.
(x−h)2+(y−k)2=r2(x-h)^2+(y-k)^2=r^2

3. Guided example: find the radius and write the equation

Suppose a circle has centre (−2,1)(-2,1) and passes through the point (1,5)(1,5). The point is on the circle, so the distance from the centre to this point is the radius. First find the horizontal and vertical changes. Then use those changes to find the distance.
The horizontal change is 1−(−2)=31-(-2)=3. The vertical change is 5−1=45-1=4. These form the legs of a right triangle, so the radius is the distance given by the Pythagorean theorem.
Now put the centre coordinates and the radius into the circle equation. Because the centre’s horizontal coordinate is −2-2, the horizontal bracket becomes x+2x+2. The vertical bracket is y−1y-1. The radius is 55, so the right side is 2525.
To sketch the graph, plot the centre first. From that point, move 55 units right, left, up, and down to locate four points on the circle. These are (3,1)(3,1), (−7,1)(-7,1), (−2,6)(-2,6), and (−2,−4)(-2,-4). Draw a smooth round curve through these points. The curve should be the same distance from the centre all the way around.
(x+2)2+(y−1)2=25(x+2)^2+(y-1)^2=25

4. Sketching and independent practice

A reliable sketch begins with the centre and radius. Plot the centre, then mark the four points that are one radius away to the left, right, above, and below. Use a compass if one is available. Otherwise, draw a smooth curve through the four points and check that it looks round, not like a square or a diamond.
If you are given an equation in circle form, compare it with the general form. The numbers in the brackets identify the centre, remembering that the signs appear opposite to the centre coordinates. The number on the right is the radius squared, so find the positive number whose square equals it to get the radius.
If you are given a centre and radius, write the equation by placing the centre values in the brackets and squaring the radius on the right. Check your work by substituting one known point on the circle. The left side should equal the right side.
Try these independently. For a circle with centre (2,−3)(2,-3) and radius 44, write its equation and name the four points directly above, below, left, and right of the centre. For a circle with centre (1,2)(1,2) passing through (4,6)(4,6), find the radius and write its equation. For an equation with right side 3636, the radius is the positive number whose square is 3636.
r2=(x−h)2+(y−k)2r^2=(x-h)^2+(y-k)^2

Worked example

Use a point on the circle to find its radius

A circle has centre (−2,1)(-2,1) and passes through (1,5)(1,5). Find its radius, write its equation, and describe how to sketch it.
  1. Find the changes
    Subtract the centre coordinates from the point coordinates. The horizontal change is 33 units and the vertical change is 44 units.
    1−(−2)=3,5−1=41-(-2)=3,\qquad 5-1=4
  2. Find the radius
    The horizontal and vertical changes are the legs of a right triangle. The distance between the centre and the point is its hypotenuse, so use the Pythagorean theorem.
    r2=32+42=25,r=5r^2=3^2+4^2=25,\qquad r=5
  3. Write the equation
    Use the centre coordinates in the brackets and put the squared radius on the right. Subtracting the negative horizontal coordinate makes the first bracket x+2x+2.
    (x+2)2+(y−1)2=25(x+2)^2+(y-1)^2=25
  4. Plan the sketch
    Plot the centre (−2,1)(-2,1). Mark points 55 units left, right, up, and down from it. Draw a smooth circle through those landmarks.
    (3,1),(−7,1),(−2,6),(−2,−4)(3,1),\quad(-7,1),\quad(-2,6),\quad(-2,-4)
Answer: The radius is 55, and the equation is (x+2)2+(y−1)2=25(x+2)^2+(y-1)^2=25.
Check: The given point (1,5)(1,5) gives (1+2)2+(5−1)2=9+16=25(1+2)^2+(5-1)^2=9+16=25, so it satisfies the equation.

Common mistakes and how to avoid them

Writing (x−2)2(x-2)^2 when the centre’s horizontal coordinate is −2-2.
Correction: Subtract the centre coordinate: x−(−2)=x+2x-(-2)=x+2. The sign inside the bracket can look opposite to the centre coordinate.
Putting the radius, rather than its square, on the right side.
Correction: The circle equation has r2r^2 on the right. If the radius is 55, use 2525.
Sketching a circle with unequal distances from the centre.
Correction: Check that the curve stays one radius from the centre in every direction. Mark the left, right, top, and bottom points first.
Treating just the four landmark points as the whole circle.
Correction: The landmarks help guide the sketch. Draw a smooth curve through them; the circle contains many more points.

Lesson summary

Check your understanding

Question 1

A circle has centre (3,−1)(3,-1) and radius 22. Which equation represents it?
  1. (x−3)2+(y+1)2=4(x-3)^2+(y+1)^2=4
  2. (x+3)2+(y−1)2=4(x+3)^2+(y-1)^2=4
  3. (x−3)2+(y+1)2=2(x-3)^2+(y+1)^2=2
  4. (x−1)2+(y+3)2=4(x-1)^2+(y+3)^2=4
Show answer and explanation
(x−3)2+(y+1)2=4(x-3)^2+(y+1)^2=4
The brackets use x−3x-3 and y−(−1)=y+1y-(-1)=y+1. The radius squared is 22=42^2=4.

Question 2

A circle’s centre is (0,0)(0,0) and a point on it is (5,12)(5,12). What is its radius?
  1. 1313
  2. 1717
  3. 119119
  4. 169169
Show answer and explanation
1313
The distance is the square root of 52+122=1695^2+12^2=169, which is 1313.

Question 3

For the circle (x+4)2+(y−2)2=9(x+4)^2+(y-2)^2=9, what are the centre and radius?
  1. Centre (−4,2)(-4,2) and radius 33
  2. Centre (4,−2)(4,-2) and radius 33
  3. Centre (−4,2)(-4,2) and radius 99
  4. Centre (4,2)(4,2) and radius 33
Show answer and explanation
Centre (−4,2)(-4,2) and radius 33
The brackets show centre coordinates (−4,2)(-4,2). Since r2=9r^2=9, the radius is 33.

Key terms

Centre
The fixed point in the middle of a circle.
Radius
The distance from the centre to any point on the circle.
Ordered pair
A pair of numbers, written (x,y)(x,y), that gives a point’s position on a coordinate grid.
Hypotenuse
The longest side of a right triangle, opposite the right angle.

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About this lesson

Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 10 Mathematics (MPM2D), study topic G9. It is a study resource, not an official curriculum publication.

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