What is the centre of a circle and its radius on the Cartesian plane whose equation is x squared plus y squared equals 12x -10y -12 showing key aspects of work?

Answer 1

Equation of the circle is:

x2 + y2 = 12x - 10y - 12

which can written as:

x2 - 12x + y2 + 10y = -12

Now by the method of completing square we can get the coordinates of the center of the circle:

Coefficient of x2 = 1

Coefficient of x = -12 = -2(6)

So -12x can be written as -2(x)(6) ...(1)

It is clear that by adding suitable term we obtain (a - b)2 or (a + b)2

The term -2ab is in the expansion of (a - b)2 so:

From 1 it is clear that b is 6.

So we need to add 62 to both sides of the equation.

Coefficient of y2 = 1

Coefficient of y = 10 = 2(5)

So 10y can be written as 2(y)(5) ...(2)

The term 2ab is in the expansion of (a + b)2 so:

From 2 it is clear that b is 5.

So we need to add 52 to both sides of the equation.

The equation of circle, now, becomes:

x2 - 12x + 62 + y2 + 10y + 52 = -12 + 62 + 52

(x - 6)2 + (y + 5)2 = 49

(x - 6)2 + (y + 5)2 = 72

(x - 6)2 + (y - (-5))2 = 72

So the coordinates of the center is 6,-5 and its radius is 7 units.