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 x^{2}+y^{2}-4x-8y-41=0 and is concentric with x^{2}+y^{2}-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* – 2*y* = 3

∴ *r* – 2*r* = 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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