Question

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

Answer 1

GOOD MORNING!!!

x² + y² -6x -4y -12 = 0

Co-ordinates of it's centre is ( 3 , 2 )

EQUATION OF REQUIRED CIRCLE IS

( x - h )² + ( y - k )² = r²

Where h = 4 And k = 5

r² = ( 3 -4 )² + ( 2 - 5 )²

r² = 1 + 9

r² = 10

So, Equation of circle is

( x - 4 )² + ( y - 5 )² = 10

OR

x² + y² -8x -10y + 31=0

Answer 2

x2+y2-6x-4y-12=0

4^2+5^2-6x-4y-12=0

16+25-6x-4y-12=0

32-12-6x-4y-12=0

-6x-4y+20=0

6x+4y=20