How do you write an equation of a circle with Center (5, -6) Point on Circle: (18 -6)?

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Answer 1 · 2016-11-23T07:13:52

The equation of the circle can be written:

#(x-5)^2+(y+6)^2 = 169#

or

#x^2+y^2-10x+12y-147 = 0#

The distance between #(5, -6)# and #(18, -6)# is #18-5 = 13#, since both lie on the horizontal line #y = -6#.

If the points did not lie on the same horizontal or vertical line then you would use the Pythagorean distance formula:

#d = sqrt((x_2-x_1)^2+(y_2-y_1)^2)#

but it would give the same answer:

#d = sqrt((18-5)^2+((-6)-(-6))^2) = sqrt(13^2+0^2) = 13#

Thus the radius of the circle is #13#.

The equation of a circle with centre #(h, k)# and radius #r# can be written:

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

So in the given example, we can write:

#(x-5)^2+(y-(-6))^2=13#

which simplifies to:

#(x-5)^2+(y+6)^2 = 169#

If you prefer, this can be expanded to:

#x^2-10x+25+y^2+12y+36=169#

then simplified to:

#x^2+y^2-10x+12y-147 = 0#