find the equation of circle whose centre is the point of intersection of the lines 2x-3y+4=0 and 3x+4y-5=0 passes through the origin.
find the equation of circle which passes through the centre of the circle x2+y2-4x-8y-41=0 and is concentric with x2+y2-2y+1=0
prove tht points (2,-4)(3,-1)(3,-3)(0, 0)are concyclic
answer fast.urgent
Q1. Let the centre and radius of the circle be (h, k) and r respectively.
∴ Equation of the circle is
(x – h)2 + (y – k)2 = r 2 ...(1)
When the circle touches both the axes, then
h = k = r
From (1), we have
(x – r)2 + (y – r)2 = r 2 ...(2)
∴ Centre of the circle = (r, r)
Given, centre of the circle lie on x – 2y = 3
∴ r – 2r = 3
⇒ – r = 3
⇒ r = – 3
The equation of the circle is
[x – (– 3)]2 + [y – (– 3)]2 = (– 3)2 (Using (2))
⇒ (x + 3)2 + (y + 3)2 = 9
Hence, the equations of the circle is (x + 3)2 + (y + 3)2 = 9.
Q3. Prove that points (2, -4), (3, -1), (3, -3), (0, 0) are concyclic.
SOL: Let us find the equation of the circle passing through the points (2, -4), (3, -1) and (3, -3).
Let the equation of this circle be
x² + y² + 2gx + 2fy + c = 0 .......(1)
As the points (2, -4), (3, -1) and (3, -3) lie on it, we get
4 + 16 + 4g -8f + c = 0 .........(2)
9 + 1 + 6g -2f + c = 0............(3)
9 + 9 + 6g - 6f + c = 0...........(4)
On solving equations (2), (3) and (4), we get
f = 2, g = -1 and c = 0.
Substituting these values in equation (1), we get
x² + y² - 2x + 4y = 0.......(5)
The fourth point (0, 0) will lie on (5) if 0 +0 - 0 - 0 = 0 i.e. if 0 = 0, which is true.
Hence, the given points are concyclic.
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