How do you find the equation of a circle passes through point (-3,6) & touch the two axis?

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Answer 1 · 2016-07-19T16:57:25

Two solutions:

#(x+15)^2+(y-15)^2=15^2#

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

The centre of the circle is in Q2 since the point #(-3, 6)# is.

So the equation of the circle may be written:

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

This is satisfied by #(-3, 6)#, so:

#(-3+r)^2+(6-r)^2 = r^2#

Expanding:

#9-6r+r^2+36-12r+r^2 = r^2#

Simplifying:

#0 = r^2-18r+45#

#= r^2-18r+81-36#

#= (r-9)^2-6^2#

#= ((r-9)-6)((r-9)+6)#

#= (r-15)(r-3)#

So #r=15# or #r=3#

graph{((x+15)^2+(y-15)^2-15^2)((x+3)^2+(y-3)^2-3^2)((x+3)^2+(y-6)^2-0.04) = 0 [-16.51, 8.8, -1.37, 11.29]}