How do you find the equation of the circle with center at (-3, 1) and through the point (2, 13)?

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Answer 1 · 2018-07-09T08:42:26

#(x-1)^2+(y+2)^2=169#

Let the equation of unknown circle with center #(x_1, y_1)\equiv(-3, 1)# & radius #r# be as follows

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

#(x-(-3))^2+(y-1)^2=r^2#

#(x+3)^2+(y-1)^2=r^2#

Since, the above circle passes through the point #(2, 13)# hence it will satisfy the equation of circle as follows

#(2+3)^2+(13-1)^2=r^2#

#r^2=25+144=169#

setting #r^2=169#, we get the equation of circle

#(x-1)^2+(y+2)^2=169#