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Nezavi [6.7K]
3 years ago
11

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

Mathematics
2 answers:
BlackZzzverrR [31]3 years ago
5 0
(x+4)^2 + (y-3)^2 = 4

Which means, radius is 2, center is (-4,3)
Novosadov [1.4K]3 years ago
4 0

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.

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