Math, asked by narendargupta9476, 1 month ago

Find the center and radius of the following circles x ^ 2 + y ^ 2 - 4x - 8y - 41 = 0, 3x ^ 2 + 3y ^ 2 + 6x - 12y - 1 = 0​

Answers

Answered by upadhyaykushagra003
0

Answer:

By ax^2+by^2+2gx+2fy+c=0

We have for the first circle,g= 2,f= -4,c= -41

and for second,g=1,f= -3,c= -1/3

by formula center is( -g,-f) and radius is,r=√g^2+f^2-c. (it's a whole square root)

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