Que- find the equation which touches the axis of x at a distance 3 from the origin and intercepts a length 6 on the axis of y.

Que-a circle has radius 3 units and its centre lies on line y=x-1.find the equation of circle if it passes through (7,3).

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If your questions are:

**Que 1:** find the equation of the circle which touches the axis of *x* at a distance 3 from the origin and intercepts a length 6 unit on the axis of *y*.

**Que 2: **A circle has radius 3 units and its centre lies on line *y* = *x *– 1. Find the equation of circle if it passes through (7,3).

**Solution 1:**

Let G and G' be the centres of the give circle. The diagram is shown below.

Here,

AB = 6 unit, OP = OQ = 3 unit,

So, MG = OP = 3 unit and MA = 1/2 (AB) = 3 unit

Using Pythagoras theorem in right triangle AGM, we have

GA^{2} = MG^{2} + MA^{2}

GA^{2} = 3^{2} + 3^{2}

GA = 3√2 unit

Also, GP = GA = 3√2 unit

Thus, the coordinates of the centres are (-3, 3√2) and (3, 3√2) and radius is 3√2 unit.

Hence, required equations will be as follows:

(*x* + 3 )^{2} + (*y* - 3√2)^{2 }= (3√2)^{2} and (*x* - 3 )^{2} + (*y* - 3√2)^{2 }= (3√2)^{2}

**Solution 2:**

Coordinates of any point lying on the line *y* = *x *– 1 will be of form (*k*, *k *– 1).

Let G (*k*, *k *– 1) be the centre of the given circle.

Radius of the circle = 3 cm (Given)

Therefore, equation of required circle

(*x* – *k*)^{2} + {*y* – (*k* – 1)}^{2} = 3^{2}

Since it passes through (7, 3), therefore

(7 – *k*)^{2} + {3 – (*k* – 1)}^{2} = 3^{2}

(7 – *k*)^{2} + (4 – *k* )^{2} = 9

*k*^{2} – 11*k* + 28 = 0

(*k* – 7)(*k* – 4) = 0

*k* = 4, 7

On substituting the value of *k *in (1), we get the equation of required circle as follows:

(*x* – 4)^{2} + {*y* – 3}^{2} = 3^{2 }and * *(*x* – 7)^{2} + {*y* – 6}^{2} = 3^{2}

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