Find the equation of a circle that passes through (2, 0) and (8, 0) and also touches the y axis.   (3 marks)

Solution

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

\(
(8 - 5)^2 + (0 - b)^2 = 5^2
\)
\(
3^2 + b^2 = 5^2
\)

\(
b^2 = 25 - 9
\)
\(
b^2 = 16
\)
\(
b = \pm4
\)

\(\text{Centre } C(5,4)\) \( \text{ or } (5,-4)\)
Equation  

 \( (x - 5)^2 + (y - 4)^2 = 5^2\)

\(\text{Or} \quad (x - 5)^2 + (y + 4)^2 = 25\)

 10. \(4x^2 + 4y^2 - 16x + 24y + 3 = 0\), find the centre of the circle and its radius. (3 marks) 

\((x - a)^2 + (y - b)^2 = r^2\) 

\(x^2 + y^2 - 4x + 6y + \frac{3}{4} = 0\) 

\(x^2 - 4x + 4 + y^2 + 6y + 9\)

\( = -\frac{3}{4} + 4 + 9\)

 \((x - 2)^2 + (y + 3)^2 \)

\(= \frac{49}{4}\) 

Centre \((2, -3) \; r = 3.5\)

The points with co-ordinates A(13,3) and B(−3,−9)   ) are the end of a diameter of a circle centre O. Determine:

(i) The coordinates of O (1 mark)

\(
\text{Midpoint} = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)
\)

\(
O = \left( \frac{13 + (-3)}{2}, \frac{3 + (-9)}{2} \right)\)
\(= \left( \frac{10}{2}, \frac{-6}{2} \right)\)

\( = (5, -3)
\)

O(5,-3)

(ii) The equation of the circle expressing it in the form \(x^2+y^2+ax+by+c=0\) (2 marks)

\(
(5 - (-13), -3 - 3) = (-8, -6)
\)

\(
|AO| = \sqrt{(-8)^2 + (-6)^2}\)

\( = \sqrt{64 + 36} = \sqrt{100} = 10
\)

\(
(x - 5)^2 + (y + 3)^2 = 10^2
\)

\(
x^2 - 10x + 25 + y^2 + 6y + 9 - 100 = 0
\)

\(
x^2 + y^2 - 10x + 6y - 66 = 0
\)

The diameter \(AB\) of a circle passes through points \(A(4,1)\) and \(B(2,1)\).
Find the equation of the circle and leave your equation in the form:
\(
x^2 + y^2 + ax + by = c
\)  where \(a\), \(b\), and \(c\) are constants.

Given points \(A(-4,1)\) and \(B(2,1)\), the center \(C\) of the circle is the midpoint of \(AB\):
\(
C\left(\frac{4+2}{2}, \frac{1+1}{2}\right) = C(-1,1)
\)

To find the radius, calculate the distance between \(A\) and \(C\):
\(
|\overline{AC}| = \sqrt{(-1 -(- 4))^2 + (1 - 1)^2}\)

\( = \sqrt{(3)^2 + 0^2} \)

\(= \sqrt{9} = 3
\)

The general form of the circle is:
\(
(x - a)^2 + (y - b)^2 = r^2
\)
Substituting \(a = -1\), \(b = 1\), and \(r = 3\):
\(
(x + 1)^2 + (y - 1)^2 = 3^2
\)
\(
(x + 1)^2 + (y - 1)^2 = 9
\)

Expanding and simplifying:
\(
(x^2 + 2x + 1) + (y^2 - 2y + 1) = 9
\)
\(
x^2 + 2x + y^2 - 2y + 2 = 9
\)
\(
x^2 + y^2 + 2x - 2y = 7
\)

This is the required equation in the form \(x^2 + y^2 + ax + by = c\), where \(a = 2\), \(b = -2\), and \(c = 7\).

A circle of radius 7cm has its centre at the point of intersection between two lines 
   \( x + 2y + 1 = 0 \) and \( 2x + 3y - 3 = 0 \). Find the equation of the circle expressing 
   it in the form \( x^2 + y^2 + gx + fy + c = 0 \).

KCSE 2006, PP2, Q10

The points with coordinates (5,5) and (-3,-1) are the ends of a diameter of a circle centre A. Determine:

(a) the coordinates of A; (1 mark)

(b) the equation of the circle expressing it in the form \( x^2 + y^2 + ax + by + c = 0 \), where \( a, b \) and \( c \) are constants. (3 marks)

Solution
(a)

\(
A \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \)

\(= \left( \frac{5 + (-3)}{2}, \frac{5 + (-1)}{2} \right)\)

\( = (1,2)
\)

∴ \( A(1,2) \)

(b)

Radius \( r \) is given by:


Using (5,5) and centre (1,2):
\(
r = \sqrt{4^2 + 3^2} = 5
\)

∴ Equation of the circle:

\(
(x - 1)^2 + (y - 2)^2 = 5^2
\)

Expanding the left-hand side:

\(
x^2 - 2x + 1 + y^2 - 4y + 4 = 25
\)

Thus,

\(
x^2 + y^2 - 2x - 4y - 20 = 0
\)

∴ \( a = -2 \), \( b = -4 \), and \( c = -20 \).