Question

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

Answer 1

C=(4,5)

centre of give circle =P(-2,3)

radius of the required circle =CP=

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

$= \sqrt{36 + 4} = \sqrt{40}$

equation of the required circle is

$(x - 4) {}^{2} + (y - 5) {}^{2} = { \sqrt{40} }^{2}$

${x}^{2} + {y}^{2} - 8x - 10y + 1 = 0$

I hope it will help you