A circle is drawn through the points (3, 2), (-1, -2) and (5, -4). Find the :
(a) coordinates of the centre of the circle ;
(b) radius of the circle ;
(c) equation of the circle.
Explanation
((x – a)^{2} + (y – b)^{2} = r^{2})
(x^{2} – 2ax + a^{2} + y^{2} – 2by + b^{2} = r^{2})
Using A(3, 2) :
(9 – 6a + a^{2} + 4 – 4b + b^{2} = r^{2})
(13 – 6a – 4b + a^{2} + b^{2} = r^{2} …. (1))
Using B(-1, -2) :
(1 + 2a + a^{2} + 4 + 4b + b^{2} = r^{2})
(5 + 2a + 4b + a^{2} + b^{2} = r^{2} …. (2))
Using C(5, -4) :
(25 – 10a + a^{2} + 16 + 8b + b^{2} = r^{2})
(41 – 10a + 8b + a^{2} + b^{2} = r^{2} … (3))
(1) – (2) :
(8 – 8a – 8b = 0 implies 1 – a – b = 0 …. (4))
(3) – (2) :
(24 – 12a + 12 + 4b = 0 implies 36 – 12a + 4b = 0 )
(equiv 9 – 3a + b = 0… (5))
Solving (4) and (5), we have :
(a = 2frac{1}{2} ; b = -1frac{1}{2})
Coordinates of the centre : (C(2frac{1}{2}, -1frac{1}{2})).
(b) To get the radius, we can make use of any point on the circle as well as the centre.
Using point (A(3, 2) & C(2frac{1}{2}, -1frac{1}{2}))
Radius = (sqrt{(3 – 2frac{1}{2})^{2} + (2 + 1frac{1}{2})^{2}})
= (sqrt{frac{1}{4} + frac{49}{4}})
= (sqrt{frac{50}{4}})
= (frac{5sqrt{2}}{2} cm)
(c) Equation of the circle :
((x – 2frac{1}{2})^{2} + (y + 1frac{1}{2})^{2} = (frac{5sqrt{2}}{2})^{2})
(x^{2} – 11x + frac{25}{4} + y^{2} + 3y + frac{9}{4} = frac{50}{4})
(x^{2} + y^{2} – 11x + 3y + frac{34}{4} – frac{50}{4} = 0)
(x^{2} + y^{2} – 11x + 3y – 4 = 0)