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Question

Question: The equation of the circle passing through the points (0, 0), (0, b) and (a, b) is....

The equation of the circle passing through the points (0, 0), (0, b) and (a, b) is.

A

x2+y2+ax+by=0x ^ { 2 } + y ^ { 2 } + a x + b y = 0

B

x2+y2ax+by=0x ^ { 2 } + y ^ { 2 } - a x + b y = 0

C

x2+y2axby=0x ^ { 2 } + y ^ { 2 } - a x - b y = 0

D

x2+y2+axby=0x ^ { 2 } + y ^ { 2 } + a x - b y = 0

Answer

x2+y2axby=0x ^ { 2 } + y ^ { 2 } - a x - b y = 0

Explanation

Solution

Equation of circle passing through (0, 0) is

x2+y2+2gx+2fy=0x ^ { 2 } + y ^ { 2 } + 2 g x + 2 f y = 0 ….(i)

Also, circle (i) is passing through (0, b) and (a, b)

\therefore f=b2f = - \frac { b } { 2 } and a2+b2+2ag+2(b2)b=0a ^ { 2 } + b ^ { 2 } + 2 a g + 2 \left( - \frac { b } { 2 } \right) b = 0

g=a2\Rightarrow g = - \frac { a } { 2 }

Hence the equations of circle is, x2+y2axby=0x ^ { 2 } + y ^ { 2 } - a x - b y = 0.