Find the equation of circle with radius 5 units and which lies within the circle #x^2+y^2+14x+10y-26=0# and which touches the given circle at point #(-1,3)#.?

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Answer 1 · 2018-06-12T12:42:36

The equation is #x^2+8x+y^2+2y-8=0#

The circle is

#x^2+y^2+14x+10y-26=0#

#x^2+14x+49+y^2+10y+25=26+49+25#

#(x+7)^2+(x+5)^2=10^2#

The center of the circle is #C(-7.-5)# and the radius is #r=10#

The new circle has a radius of #r_1=5#

The center of the new circle lies on the mid-point of #(-7,-5)# and #(-1,3)#

The center is #C_1=((-7-1)/2, (-5+3)/2)=(-4,-1)#

The equation of the new circle is

#(x+4)^2+(y+1)^2=25#

#x^2+8x+y^2+2y-8=0#

graph{(x^2+14x+y^2+10y-26)(x^2+8x+y^2+2y-8)=0 [-27.49, 18.14, -14.63, 8.18]}