Intersection of equation of circle x²+y²-2x-6y+1=0 and x²+y²-10x-12y+47=0?

1 Answer
Shwetank Mauria
Mar 7, 2018

Points of intersection are #(3.942,2.411)# and #(1.258,5.989)#

Explanation:

To find points of intersection of circles #x^2+y^2-2x-6y+1=0# and #x^2+y^2-10x-12y+47=0#, we should solve the two equations as simultaneous equations.

Subtracting latter equation from former, we get

#8x+6y-46=0# or #4x+3y-23=0# or #y=-4/3x+23/3#

Now putting this value of #y# in first equation, we get

#x^2+(-4/3x+23/3)^2-2x-6(-4/3x+23/3)+1=0#

or #x^2+16/9x^2-184/9x+529/9-2x+8x-46+1=0#

or #9x^2+16x^2-184x+529+54x-405=0#

or #25x^2-130x+124=0#

and using quadratic formula

#x=(130+-sqrt(130^2-12400))/50#

= #(130+-sqrt4500)/50=13/5+-3/5sqrt5#

if #x=13/5+3/5sqrt5=3.942# and

#y=-4/3(13/5+3/5sqrt5)+23/3=-52/15-4/5sqrt5+23/3=2.411#

and if #x=13/5-3/5sqrt5=1.258# and

#y=-4/3(13/5-3/5sqrt5)+23/3=-52/15+4/5sqrt5+23/3=5.989#

Hence, points of intersection are #(3.942,2.411)# and #(1.258,5.989)#

graph{(x^2+y^2-2x-6y+1)(x^2+y^2-10x-12y+47)=0 [-7.71, 12.29, -0.44, 9.56]}

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