Question

Find the equation of the circle for which the points given below are the end points of a diameter (-4,3),(3,_4)​

Answer 1

Answer:

(2x + 1)² + (2y + 1)² = 98

Step-by-step explanation:

The standard form for the equation of a circle is (x - h)² + (y - k)² = r² , where (h, k) are coordinates of a center and "r" is a radius.

( $\frac{x_{1} +x_{2} }{2}$ , $\frac{y_{1} +y_{2} }{2}$ ) - coordinates of midpoint

d = $\sqrt{(x_{2} -x_{1} )^2 + (y_{2} -y_{1} )^2}$ - distance between two points.

~~~~~~~~~~~~~~

If ( - 4, 3) and (3, - 4) are the end points of a diameter, then midpoint will be center of the circle.

( $\frac{-4+3}{2}$ , $\frac{3-4}{2}$ ) = ( - $\frac{1}{2}$ , - $\frac{1}{2}$ )

r = $\frac{\sqrt{(-4-3)^2 +(3+4)^2} }{2}$ = $\frac{7\sqrt{2} }{2}$

(x + $\frac{1}{2}$ )² + (y + $\frac{1}{2}$ )² = $\frac{98}{4}$ ⇔ (2x + 1)² + (2y + 1)² = 98