# Textbooks for High School Students Studying the Mathematics

## 42.3 Co-ordinate Geometry

**42.3.1 Equation of a Circle
**We know that every point on the circumference of a circle is the same distance
away from the

centre of the circle. Consider a point (x

_{1},y

_{1}) on the circumference of a circle of radius r with

centre at (x

_{0},y

_{0}).

Figure 42.3: Circle h with centre (x

_{0},y

_{0}) has a tangent, g passing through point P at (x

_{1},y

_{1}).

Line f passes through the centre and point P.

In Figure 42.3, △OPQ is a right-angled triangle. Therefore, from the Theorem of
Pythagoras,

we know that:

OP^{2} = PQ^{2} + OQ^{2}

But,

PQ = y_{1} - y_{0}

OQ = x_{1} - x_{0}

OP = r

r^{2} = (y_{1} - y_{0})^{2} + (x_{1} - x_{0})^{2}

But, this same relation holds for any point P on the circumference. In fact, the
relation holds

for all points P on the circumference. Therefore, we can write:

(x - x_{0})^{2} + (y - y_{0})^{2} = r^{2} (42.3)

for a circle with centre at (x_{0},y_{0}) and radius r.

For example, the equation of a circle with centre (0,0) and radius 4 is:

(y - y_{0})^{2} + (x - x_{0})^{2} = r^{2}

(y - 0)^{2} + (x - 0)^{2} = 4^{2}

y^{2} + x^{2} = 16

**Worked Example 192: Equation of a Circle I**

**Question:** Find the equation of a circle (centre O) with a diameter between two

points, P at (-5,5) and Q at (5, - 5).

**Answer
Step 1 : Draw a picture
**Draw a picture of the situation to help you figure out what needs to be done.

**Step 2 : Find the centre of the circle**

We know that the centre of a circle lies on the midpoint of a diameter.
Therefore

the co-ordinates of the centre of the circle is found by finding the midpoint of
the

line between P and Q. Let the co-ordinates of the centre of the circle be
(x_{0},y_{0}),

let the co-ordinates of P be (x_{1},y_{1}) and let the co-ordinates of Q be (x_{2},y_{2}).
Then,

the co-ordinates of the midpoint are:

The centre point of line PQ and therefore the centre of the circle is at (0,0).

**Step 3 : Find the radius of the circle
**If P and Q are two points on a diameter, then the radius is half the distance
between

them.

The distance between the two points is:

**Step 4 : Write the equation of the circle**

x

^{2}+ y

^{2}= 50

**Worked Example 193: Equation of a Circle II**

**Question:** Find the center and radius of the circle

x^{2} - 14x + y^{2} + 4y = -28.

**Answer
Step 1 : Change to standard form
**We need to rewrite the equation in the form (x - x

_{0}) + (y - y

_{0}) = r

^{2}

To do this we need to complete the square

i.e. add and subtract ( 1/2 cooefficient of x)

^{2}and ( 1/2 cooefficient of y)

^{2}

**Step 2 : Adding cooefficients
**x

^{2}- 14x + y

^{2}+ 4y = -28

x

^{2}- 14x + (7)

^{2}- (7)

^{2}+ y

^{2}+ 4y + (2)

^{2}- (2)

^{2}= -28

**Step 3 : Complete the squares**

(x - 7)

^{2}- (7)

^{2}+ (y + 2)

^{2}- (2)

^{2}= -28

**Step 4 : Take the constants to the other side**

(x - 7)

^{2}- 49 + (y + 2)

^{2}- 4 = -28

(x - 7)

^{2}+ (y + 2)

^{2}= -28 + 49 + 4

(x - 7)

^{2}+ (y + 2)

^{2}= 25

**Step 5 : Read the values from the equation**

center is (7;-2) and the radius is 5 units

**42.3.2 Equation of a Tangent to a Circle at a Point on the Circle
**We are given that a tangent to a circle is drawn through a point P with
co-ordinates (x

_{1},y

_{1}).

In this section, we find out how to determine the equation of that tangent.

Figure 42.4: Circle h with centre (x

_{0},y

_{0}) has a tangent, g passing through point P at (x

_{1},y

_{1}).

Line f passes through the centre and point P.

We start by making a list of what we know:

1. We know that the equation of the circle with centre (x_{0},y_{0}) is (x-x_{0})^{2} +(y
-y_{0})^{2} = r^{2}.

2. We know that a tangent is perpendicular to the radius, drawn at the point of
contact with

the circle.

As we have seen in earlier grades, there are two steps to determining the
equation of a straight

line:

Step 1: Calculate the gradient of the line, m.

Step 2: Calculate the y-intercept of the line, c.

The same method is used to determine the equation of the tangent. First we need
to find the

gradient of the tangent. We do this by finding the gradient of the line that
passes through the

centre of the circle and point P (line f in Figure 42.4), because this line is a
radius line and the

tangent is perpendicular to it.

The tangent (line g) is perpendicular to this line. Therefore,

So,

Now, we know that the tangent passes through (x_{1},y_{1}) so the equation is given
by:

For example, find the equation of the tangent to the circle at point (1,1). The
centre of the

circle is at (0,0). The equation of the circle is x^{2} + y^{2} = 2.

Use

with (x_{0},y_{0}) = (0,0) and (x_{1},y_{1}) = (1,1).