Question

Find the equation of circle the endpoints of whose diameter are (2,-3) and (-2,4)​

Answer 1

Given ,

The endpoints of diameter of circle are (2,-3) and (-2,4)

We know that ,

if $\tt(x_{1} , y_{1})$ and$\tt(x_{2} , y_{2})$ are the ends points of diameter of circle , then the equation of circle is given by

$\boxed{ \tt{(x - x_{1}) (x - x_{2}) + (y- y_{1}) (y - y_{2}) = 0}}$

Thus ,

$\tt \hookrightarrow \{x - 2 \} \{ x - ( - 2)\} + \{y - ( - 3) \} \{y - 4 \} = 0$

$\tt \hookrightarrow\{ x - 2\} \{ x + 2\} + \{ y + 3\} \{y - 4 \} = 0$

$\tt \hookrightarrow{(x)}^{2} - {(2)}^{2} + {(y)}^{2} - 4y + 3y - 12 = 0$

$\tt \hookrightarrow {(x)}^{2} - 4 + {(y)}^{2} - y - 12 = 0$

$\tt \hookrightarrow{(x)}^{2} + {(y)}^{2} - y - 16 = 0$

The required equation of circle is

• (x)² + (y)² - y - 16 = 0

Answer 2

Answer:

(x)² + (y)² - y - 16 = 0

is the answer.