The equation of a circle is #3x^2 + 3y^2 -2x +my - 2 = 0#. What is the value of m if the point (4,3) lies on the circle?

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Answer 1 · 2016-07-28T20:25:16

#m=-65/3#

Substitute #x=4#, #y=3# into the equation to find:

#3(4^2)+3(3^2)-2(4)+m(3)-2 = 0#

That is:

#48+27-8+3m-2 = 0#

That is:

#3m+65 = 0#

So #m = -65/3#

graph{(3x^2+3y^2-2x-65/3y-2)((x-4)^2+(y-3)^2-0.02) = 0 [-8.46, 11.54, -2.24, 7.76]}