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Question
Find the equation of the circle with centre at (−3, −3) passing through the point (−3, −6)
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Solution
Centre of the circle is C (– 3, – 3) and it passes through the point P (– 3, – 6).
By distance formula
Radius (r) = CP = `sqrt([-3 - (-3)]^2 + [- 6 - (- 3)]^2`
= `sqrt((-3 + 3)^2 + (-6 + 3)^2`
= `sqrt(0^2 + (-3)^2`
= `sqrt(9)`
= 3
The equation of a circle with centre at (h, k) and radius ‘r’ is given by
(x – h)2 + (y – k)2 = r2
Here, h = – 3, k = –3, r = 3
∴ The required equation of the circle is
[x – (– 3)]2 + [y – (– 3)]2 = 32
∴ (x + 3)2 + (y + 3)2 = 9
∴ x2 + 6x + 9 + y2 + 6y + 9 – 9 = 0
∴ x2 + y2 + 6x + 6y + 9 = 0.
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