Find the coordinates of the centre of the circle passing through the points P(6,-6), Q(3,-7)and R(3,3).
20. Find the coordinates of the centre of the circle passing through the points P(6,-6), Q(3,-7)and R(3,3).
Solution
Let O(a, b) be the centre of the circle.
Let the points (6,- 6), (3, -7), and (3, 3) represent the points P, Q, and R on the circumference of the circle.
Distance from centre O to P, Q, R are found below using the Distance formula.
Distance Formula = `sqrt((x_2 – x_1)^2 + (y_2 – y_1)^2)`
From the figure ,
OP = OQ …(radii of the same circle)
`∴ sqrt((a – 6)^2 + (b – (- 6))^2) = sqrt((a – 3)^2 + (b – (- 7))^2)`
`∴ sqrt((a – 6)^2 + (b + 6)^2) = sqrt((a – 3)^2 + (b + 7)^2)`
Squaring on both sides,
(a – 6)2 + (b + 6)2 = (a – 3)2 + (b + 7)2
∴ a2 – 12a + 36 + b2 + 12b + 36 = a2 – 6a + 9 + b2 + 14b + 49
∴ 3a + b = 7 …(1)
OP = OR …(radii of the same circle)
`∴ sqrt((a – 6)^2 + (b – (- 6))^2) = sqrt((a – 3)^2 + (b – 3)^2)`
`∴ sqrt((a – 6)^2 + (b + 6)^2) = sqrt((a – 3)^2 + (b – 3)^2)`
Squaring on both sides,
(a – 6)2 + (b + 6)2 = (a – 3)2 + (b – 3)2
∴ a2 – 12a + 36 + b2 + 12b + 36 = a2 – 6a + 9 + b2 – 6b + 9
54 = 6a + 18
∴ a – 3b = 9 …(2)
Multiplying (2) with 3, we get,
∴ 3a – 9b = 27 …(3)
Subtracting equation (3) from (1),
[begin{array}{l}
phantom{texttt{0}}texttt{3a + b = 7}\ phantom{texttt{}}texttt{-3a – 9b = 27}\ hlinephantom{texttt{}}texttt{(-) (+) (-)}\ hline end{array}]
∴ 10b = – 20
∴ b = – 2
Substituting b = – 2 in equation (1),
3a + b = 7
3a – 2 = 7
3a = 7 + 2
3a = 9
a = 3
Coordinates of centre of the circle are (3, -2) .
Explanation:-
We are given three points, P(6,-6), Q(3,-7), and R(3,3) on the circumference of a circle. We are asked to find the coordinates of the center of the circle.
Let the center of the circle be denoted by O(a, b). Since P, Q, and R are points on the circumference of the same circle, their distances from the center O are equal.
We use the distance formula to calculate the distances between the points and the center:
Distance Formula: √((x2 – x1)^2 + (y2 – y1)^2)
- OP = OQ (radii of the same circle) ∴ √((a – 6)^2 + (b + 6)^2) = √((a – 3)^2 + (b + 7)^2)
Squaring both sides, we get:
(a – 6)^2 + (b + 6)^2 = (a – 3)^2 + (b + 7)^2
Expanding and simplifying, we get:
3a + b = 7 ………..(1)
- OP = OR (radii of the same circle) ∴ √((a – 6)^2 + (b + 6)^2) = √((a – 3)^2 + (b – 3)^2)
Squaring both sides, we get:
(a – 6)^2 + (b + 6)^2 = (a – 3)^2 + (b – 3)^2
Expanding and simplifying, we get:
a – 3b = 9 ………..(2)
Multiplying equation (2) by 3, we get:
3a – 9b = 27 ………..(3)
Subtracting equation (3) from equation (1), we get:
10b = -20
∴ b = -2
Substituting b = -2 in equation (1), we get:
3a + b = 7
∴ 3a – 2 = 7
∴ 3a = 9
∴ a = 3
Therefore, the coordinates of the center of the circle are (3, -2).
Chapter 5. Co-ordinate Geometry – Problem set 5 (Page 122)