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

GA2 = MG2 + MA2

GA2 = 32 + 32

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

(xk)2 + {y – (k – 1)}2 = 32

Since it passes through (7, 3), therefore

(7 – k)2 + {3 – (k – 1)}2 = 32

(7 – k)2 + (4 – k )2 = 9

k2 – 11k + 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 = 32  and (x – 7)2 + {y – 6}2 = 32

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x/3+y/6=1

from the general equation x/a+y/b=1  (where a is a intercept on x axis and b is intercept on y axis)  

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