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} = *r*^{2} ...(1)

The line 5*x* + 12*y* – 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

(*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.

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