Q. 133.7( 110 Votes )

# Find the eq

Let the equation of the required circle be (x – h)2+ (y – k)2 =r2

Since, the circle passes through (0, 0),

(0 – h)2+ (0 – k)2 =r2

h2 + k2 = r2

The equation of the circle now becomes (x – h)2+ (y – k)2 = h2 + k2.

It is given that the circle makes intercepts a and b on the coordinate axes.

That means that the circle passes through points (a, 0) and (0,b). Therefore,

(a – h)2+ (0 – k)2 =h2 +k2.................(1)

(0 – h)2+ (b– k)2 =h2 +k2..................(2)

From equation (1), we obtain

a2 – 2ah + h2 +k2 = h2 +k2

a2 – 2ah = 0

a(a – 2h) =0

a = 0 or (a -2h) = 0

However, a 0; hence, (a -2h) = 0

h =.

From equation (2), we obtain

h2  – 2bk + k2 + b2= h2 +k2

b2 – 2bk = 0

b(b– 2k) = 0

b= 0 or (b-2k) =0

However, a 0; hence, (b -2k) = 0

k =.

+=

4x2 -4ax + a2 +4y2 - 4by + b2 =a2 + b2

4x2 + 4y2 -4ax – 4by =0

4( x2 +y2 -7x + 5y – 14 ) = 0

x2+y2- ax - by=0

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