Question

Find the equation of the circle whose centre is at the point (4,5) and which passes through the centre of the circle. x²+y²-6x+4y-12=0.​

Answer 1

Answer:

(x - 4)² + (y - 5)² = 50

Step-by-step explanation:

Equation of the circle is (x - a)² + (y - b)² = r² , where (a, b) are coordinates of a circle, and "r" is a radius.

Distance between two points

$d=\sqrt{(x_{2}-x_{1})^{2} + (y_{2} - y_{1})^{2}}$

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

x² + y² - 6x + 4y - 12 = 0

(x² - 6x + 9) + (y² + 4y + 4) - 9 - 4 - 12 = 0

(x - 3)² + (y + 2)² = 5²

Center of given circle is (3, - 2) and r = 5

Distance between (4, 5) and (3, -2) is

$d=\sqrt{(4-3)^{2} +(5+2)^{2}} = \sqrt{50} =5\sqrt{2}$

The coordinates of the center of a circle are (4, 5) and r = 5√2

(x - 4)² + (y - 5)² = 50 is equation of the circle.

or x² + y² - 8x - 5y - 9 = 0