Question

Find the equation of the circle whose centre is at (4,5) and which passes through the centre of the circle Xsquare +Ysquare +4x -6y-12=0

Answer 1

Answer:

x² + y² - 8x - 10y + 1 = 0

Step-by-step explanation:

Centre of second circle (a point on the required circle)

x² + y² + 4x - 6y - 12 = 0

=> ( x +2 )² + ( y - 3 ) ² = 12 + 2² + 3²

=> the centre is at ( -2, 3 )

So our circle must pass through ( -2, 3 ).

Radius of the required circle

The centre is to be at ( 4, 5 ) and it passes through ( -2, 3 ).

The radius r is then the distance between these points, so:

r² = ( 4 - -2 )² + ( 5 - 3 )² = 6² + 2² = 36 + 4 = 40.

Equation of the circle

As the centre is at ( 4, 5 ) and the radius r satisfies r² = 40, the equation of the circle is

( x - 4 )² + ( y - 5 )² = 40

=> x² - 8x + 16 + y² - 10y + 25 - 40 = 0

=> x² + y² - 8x - 10y + 1 = 0

Answer 2

Answer:

As center is (4;5) and circumference is x^2+ y^2 +4x-6y=12

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

x=-2

y=3

and radius can be find out by( 4-(-)2)^2+ (5-3) ^2

6^+2^

36+4

40

now equation will be x square + y square - 8 x minus 10 by + 1 =zero

hope it helps you