What is the standard form of the equation of a circle with a center(1,-2) and passes through (6,-6)?

2 Answers
Jul 9, 2018

The circle equation in standard form is

#(x-x_0)^2+(y-y_0)^2=r^2#

Where #(x_0,y_0); r# are the center coordinates and radius

We know that #(x_0,y_0)=(1,-2)#, then

#(x-1)^2+(y+2)^2=r^2#.

But we know that passes trough #(6,-6)#, then

#(6-1)^2+(-6+2)^2=r^2#

#5^2+(-4)^2=41=r^2#, So #r=sqrt41#

Finally we have the standard form of this circle

#(x-1)^2+(y+2)^2=41#.

Answer:

#(x-1)^2+(y+2)^2=41#

Explanation:

Let the equation of unknown circle with center #(x_1, y_1)\equiv(1, -2)# & radius #r# be as follows

#(x-x_1)^2+(y-y_1)^2=r^2#

#(x-1)^2+(y-(-2))^2=r^2#

#(x-1)^2+(y+2)^2=r^2#

Since, the above circle passes through the point #(6, -6)# hence it will satisfy the equation of circle as follows

#(6-1)^2+(-6+2)^2=r^2#

#r^2=25+16=41#

setting #r^2=41#, we get the equation of circle

#(x-1)^2+(y+2)^2=41#