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likoan [24]
3 years ago
5

The general form of the equation of a circle is x^2+y^2−8x+6y+21=0.

Mathematics
2 answers:
densk [106]3 years ago
7 0

Answer:

The coordinates of the center of the circle are (4,-3).

Step-by-step explanation:

The general form of the equation of a circle is

x^2+y^2-8x+6y+21=0

It can be written as

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

Add and subtract (\frac{-b}{2a})^2 in each parenthesis, to make perfect squares.

For first parenthesis,

(\frac{-b}{2a})^2=(frac{8}{2(1)})^2=(4)^2=16

For second parenthesis,

(\frac{-b}{2a})^2=(frac{-6}{2(1)})^2=(-3)^2=9

(x^2-8x+16-16)+(y^2+6y+9-9)+21=0

(x^2-8x+16)+(y^2+6y+9)+21-16-9=0

(x-4)^2+(y+3)^2-4=0

(x-4)^2+(y+3)^2=4                  .... (1)

The standard form of a circle is

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

Where, (h,k) is center of the circle and r is radius.

On comparing (1) and (2) we get,

h=4,k=-3,r=2

Therefore the coordinates of the center of the circle are (4,-3).

Minchanka [31]3 years ago
5 0
The anwser is (4,-3)
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