Q. 133.8( 89 Votes )

# Find the equation of the circle passing through (0,0) and making intercepts a and b on the coordinate axes.

Answer :

Let the equation of the required circle be (x – h)^{2}+ (y – k)^{2} =r^{2}

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

(0 – h)^{2}+ (0 – k)^{2} =r^{2}

h^{2} + k^{2} = r^{2}

The equation of the circle now becomes (x – h)^{2}+ (y – k)^{2} = h^{2} + k^{2}.

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} =h^{2} +k^{2}.................(1)

(0 – h)^{2}+ (b– k)^{2} =h^{2} +k^{2}..................(2)

From equation (1), we obtain

a^{2} – 2ah + h^{2} +k^{2} = h^{2} +k^{2}

a^{2} – 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

h^{2} – 2bk + k^{2} + b^{2}= h^{2} +k^{2}

b^{2} – 2bk = 0

b(b– 2k) = 0

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

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

k =.

+=

⇒ 4x^{2} -4ax + a^{2} +4y^{2} - 4by + b^{2} =a^{2} + b^{2}

⇒ 4x^{2} + 4y^{2} -4ax – 4by =0

⇒ 4( x^{2} +y^{2} -7x + 5y – 14 ) = 0

⇒ x^{2}+y^{2}- a*x* - by=0

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