Find the equation to the circle which passes through the origin and cuts off intercepts equal to $3$ and $4$ from the axes.

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Solution

A general equation of the circle having center $(- g, - f)$ and radius $\sqrt{g^{2} + f^{2} - c}$ looks like $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$
Since this passes through the origin, $c = 0$
Case 1: x intercept $= 3$ , y intercept $= 4$
This case has 4 subcases, namely
a. The circle passes through $(- 3, 0)$ and $(0, 4)$
Equation becomes $x^{2} + y^{2} + 3 x - 4 y = 0$
b. The circle passes through $(- 3, 0)$ and $(0, - 4)$
Equation becomes $x^{2} + y^{2} + 3 x + 4 y = 0$
c. The circle passes through $(3, 0)$ and $(0, 4)$
Equation becomes $x^{2} + y^{2} - 3 x - 4 y = 0$
d. The circle passes through $(3, 0)$ and $(0, - 4)$
Equation becomes $x^{2} + y^{2} - 3 x + 4 y = 0$
Case 2: x intercept $= 4$ , y intercept $= 3$
This case also has 4 subcases, as below:
a. The circle passes through $(- 4, 0)$ and $(0, 3)$
Equation becomes $x^{2} + y^{2} + 4 x - 3 y = 0$
b. The circle passes through $(- 4, 0)$ and $(0, - 3)$
Equation becomes $x^{2} + y^{2} + 4 x + 3 y = 0$
c. The circle passes through $(4, 0)$ and $(0, 3)$
Equation becomes $x^{2} + y^{2} - 4 x - 3 y = 0$
d. The circle passes through $(4, 0)$ and $(0, - 3)$
Equation becomes $x^{2} + y^{2} - 4 x + 3 y = 0$

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