Question

What are the center and radius of the circle defined by the equation x^2 + y^2 - 6x + 8y + 21=0

Answer 1

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Answer 2

Answer: The center of the circle is (3, -4) and radius is 2 units.

Step-by-step explanation: We are given to find the center and radius of the circle defined by the following equation :

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

The standard equation of a circle with center (h, k) and radius r units is given by

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

From equation (i), we have

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

Comparing the above equation with the standard equation of a circle, we get

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

Thus, the center of the circle is (3, -4) and radius is 2 units.