Question

The equation of the circle passing through the origin and cutting intercepts of length 3 and 4 units from the positive axes, is.

• A. $x ^ { 2 } + y ^ { 2 } + 6 x + 8 y + 1 = 0$
• B. $x ^ { 2 } + y ^ { 2 } - 6 x - 8 y = 0$
• C. $x ^ { 2 } + y ^ { 2 } + 3 x + 4 y = 0$
• D. $x ^ { 2 } + y ^ { 2 } - 3 x - 4 y = 0$

Answer 1

Answer: (D)

$x ^ { 2 } + y ^ { 2 } - 3 x - 4 y = 0$

Answer 2

Obviously the centre of the circle is $\left( \frac { 3 } { 2 } , 2 \right)$.

Therefore, the equation of circle is

$\left( x - \frac { 3 } { 2 } \right) ^ { 2 } + ( y - 2 ) ^ { 2 } = \left( \frac { 5 } { 2 } \right) ^ { 2 } \Rightarrow x ^ { 2 } + y ^ { 2 } - 3 x - 4 y = 0$.