Question

What is the radius of a circle whose equation is x2 + y2 + 8x – 6y + 21 = 0?

Answer 1

Answer:  The required center of the given circle is 2 units.

Step-by-step explanation:  We are given to find the radius of a circle with the following equation:

$x^2+y^2+8x-6y+21=0~~~~~~~~~~~~~~~~~~~~~~~(i)$

The standard equation of a CIRCLE with radius 'r' units and center at the point (h, k) is given by

$(x-h)^2+(y-k)^2=r^2.$

From equation (i), we have

$x^2+y^2+8x-6y+21=0\\\\\Rightarrow (x^2+8x+16)+(y^2-6y+9)-16-9+21=0\\\\\Rightarrow (x^2+2\times x\times 4+4^2)+(y^2-2\times x\times 3+3^2)-4=0\\\\\Rightarrow (x+4)^2+(y-3)^2=4\\\\\Rightarrow (x-(-4))^2+(y-3)^2=2^2.$

Comparing the above equation with the standard equation (i), we get

radius, r = 2 units  and  center, (h, k) = (-4, 3).

Thus, the required center of the given circle is 2 units.

Answer 2

Radius=2
center(-2,4)