Q. 245.0( 1 Vote )

# Equation of the circle through origin which cuts intercepts of length a and b on axes isA. x2 + y2 + ax + by = 0B. x2 + y2 - ax - by = 0C. x2 + y2 + bx + ay = 0D. none of these

Given that we need to find the equation of the circle passing through origin and cuts off intercepts a and b from x and y - axes.

Since the circle is having intercept a from x - axis the circle must pass through (a,0) and (- a,0) as it already passes through the origin.

Since the circle is having intercept b from x - axis the circle must pass through (0,b) and (0, - b) as it already passes through the origin.

Let us assume the circle passing through the points O(0,0), A(a,0) and B(0,b).

We know that the standard form of the equation of the circle is given by:

x2 + y2 + 2fx + 2gy + c = 0 ..... (1)

Substituting O(0,0) in (1), we get,

02 + 02 + 2f(0) + 2g(0) + c = 0

c = 0 ..... (2)

Substituting A(a,0) in (1), we get,

a2 + 02 + 2f(a) + 2g(0) + c = 0

a2 + 2fa + c = 0 ..... (3)

Substituting B(0,b) in (1), we get,

02 + b2 + 2f(0) + 2g(b) + c = 0

b2 + 2gb + c = 0 ..... (4)

On solving (2), (3) and (4) we get, Substituting these values in (1), we get x2 + y2 - ax - by = 0

Similarly, we get the equation x2 + y2 + ax + by = 0 for the circle passing through the points (0,0), (- a,0), (0, - b).

The equations of the circles are x2 + y2±ax±by = 0.

The correct options are (a) and (b).

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