# 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=0prove tht points (2,-4)(3,-1)(3,-3)(0, 0)are concyclicanswer fast.urgent

Q1. Let the centre and radius of the circle be (h, k) and r respectively.

∴ Equation of the circle is

(xh)2 + (yk)2 = r 2 ...(1)

When the circle touches both the axes, then

h = k = r

From (1), we have

(xr)2 + (yr)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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