Question

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

Answer 1

(x+4)^2 + (y-3)^2 = 4

Which means, radius is 2, center is (-4,3)

Answer 2

Answer:

The radius of circle whose equation is $x^2+y^2+8x-6y+21=0$ is 2 units.

Step-by-step explanation:

Given equation of  a circle is $x^2+y^2+8x-6y+21=0$

We have to find the radius of the circle whose equation is given.

Consider the given equation of  a circle is $x^2+y^2+8x-6y+21=0$

General equation of circle having center at (h,k)  with radius r is given as ,

$(x-h)^2+(y-k)^2=r^2$ ........(1)

we first write the given equation in the general form by making perfect squares,

We know $(a-b)^2=a^2+b^2-2ab$

For making perfect square of x, we have terms, $x^2+8x=x^2+2\cdot 4 \cdot x$ we just need $b^2$

On comparing w have b = 4 , thus $b^2=16$

So adding both side 16 , equation becomes,

$x^2+y^2+8x-6y+21+16=16$

$\Rightarrow (x-(-4))^2+y^2-6y+21=16$

For making perfect square of y, we have terms, $y^2-6y=y^2-2\cdot 3 \cdot y$ we just need $b^2$

On comparing w have b = 3 , thus $b^2=9$

So adding both side 9 , above equation becomes,

$\Rightarrow (x-(-4))^2+y^2-6y+21+9=16+9$

On solving, we get,

$\Rightarrow (x-(-4))^2+(y-3)^2+21=25$

$\Rightarrow (x-(-4))^2+(y-3)^2=25-21$

$\Rightarrow (x-(-4))^2+(y-3)^2=4$

$\Rightarrow (x-(-4))^2+(y-3)^2=2^2$ .....(2)

Comparing (1) and (2) , we have  r= 2

Thus, the radius of circle whose equation is $x^2+y^2+8x-6y+21=0$ is 2 units.