Answer:
The radius of circle whose equation is [tex]x^2+y^2+8x-6y+21=0[/tex] is 2 units.
Step-by-step explanation:
Given equation of a circle is [tex]x^2+y^2+8x-6y+21=0[/tex]
We have to find the radius of the circle whose equation is given.
Consider the given equation of a circle is [tex]x^2+y^2+8x-6y+21=0[/tex]
General equation of circle having center at (h,k) with radius r is given as ,
[tex](x-h)^2+(y-k)^2=r^2[/tex] ........(1)
we first write the given equation in the general form by making perfect squares,
We know [tex](a-b)^2=a^2+b^2-2ab[/tex]
For making perfect square of x, we have terms, [tex]x^2+8x=x^2+2\cdot 4 \cdot x[/tex] we just need [tex]b^2[/tex]
On comparing w have b = 4 , thus [tex]b^2=16[/tex]
So adding both side 16 , equation becomes,
[tex]x^2+y^2+8x-6y+21+16=16[/tex]
[tex]\Rightarrow (x-(-4))^2+y^2-6y+21=16[/tex]
For making perfect square of y, we have terms, [tex]y^2-6y=y^2-2\cdot 3 \cdot y[/tex] we just need [tex]b^2[/tex]
On comparing w have b = 3 , thus [tex]b^2=9[/tex]
So adding both side 9 , above equation becomes,
[tex]\Rightarrow (x-(-4))^2+y^2-6y+21+9=16+9[/tex]
On solving, we get,
[tex]\Rightarrow (x-(-4))^2+(y-3)^2+21=25[/tex]
[tex]\Rightarrow (x-(-4))^2+(y-3)^2=25-21[/tex]
[tex]\Rightarrow (x-(-4))^2+(y-3)^2=4[/tex]
[tex]\Rightarrow (x-(-4))^2+(y-3)^2=2^2[/tex] .....(2)
Comparing (1) and (2) , we have r= 2
Thus, the radius of circle whose equation is [tex]x^2+y^2+8x-6y+21=0[/tex] is 2 units.
