How do you find the equation of a circle passes through point (-3,6) & touch the two axis?
1 Answer
Jul 19, 2016
Two solutions:
#(x+15)^2+(y-15)^2=15^2#
#(x+3)^2+(y-3)^2=3^2#
Explanation:
The centre of the circle is in Q2 since the point
So the equation of the circle may be written:
#(x+r)^2+(y-r)^2 = r^2#
This is satisfied by
#(-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
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]}