# Find the equation of the circle which has its centre at the point (3,4) and touches the straight the line 5x+12y-1=0.Find the equqtion of the circle which touches the axes and whose centre lies on x-2y=3.A circle whose centre is the point of the intersection of the lines 2x-3y+4=0 and 3x+4y-5=0 passes through the origin. find its equation.A circle of radius 4 units touches the coordinate axes in the first quadrant. Find the equation of its images with respect to the line mirrors x=0 and y=0.One diameter of the circle circumscribing the rectangle ABCD is 4y=x+7. If the coordinates of A and B are (-3,4) and (5,4) respectively. Find its equation.

1.

Let the radius of the circle be r.

Centre of the circle = (3, 4).

∴ Equation of the circle is

(x – 3)2 + (y – 4)2 = r2     ...(1)

The line 5x + 12y – 1 = 0 touches the circle,

∴ Radius of the circle = Length of perpendicular from (3, 4) to the given line Hence, the equations of the circle is 2. Let the centre and radius of the circle be (h, k) and r respectively.

∴ Equation of the circle is

(xh)2 + (yk)2 = r2  ...(1)

When the circle touches both the axes, then

h = k = r

From (1), we have

(xr)2 + (yr)2 = r2  ...(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.

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