A circle has a center that falls on the line #y = 7/9x +7 # and passes through # ( 4 ,5 )# and #(8 ,7 )#. What is the equation of the circle?

Answer 1

Equation of circle is #25x^2+25y^2-198x-504y+2287=0#

The center will be at the point of intersection of line #y=7/9x+7# and perpendicular bisector of segment joining #(4,5)# and #(8,7)#, which is nothing but locus of a point that moves so that it is equidistant from these two points. Hence, its equation will be

#(x-4)^2+(y-5)^2= (x-8)^2+(y-7)^2# or

#x^2-8x+16+y^2-10y+25=x^2-16x+64+y^2-14y+49# or

#8x+4y+41-113=0# or #8x+4y-72=0# or #2x+y-18=0#

To find its intersection with #y=7/9x+7#, let us put this value of #y# in perpendicular bisector, which gives

#2x+7/9x+7-18=0# or #25/9x=11#

Hence #x=99/25# and hence #y=7/9xx99/25+7=77/25+7=252/25#.

The center is hence #(99/25,252/25)# and radius will be its distance with either of the points #(4,5)# or #(8,7)#. Let us choose the first. Hence equation of circle would be

#(x-99/25)^2+(y-252/25)^2=(99/25-4)^2+(252/25-5)^2# or

#(25x-99)^2+(25y-252)^2=(99-100)^2+(252-125)^2# (multiplying each side by #25^2#)

or #625x^2-4950x+9801+625y^2-12600x+63504=(-1)^2+127^2#

or #625x^2+625y^2-4950x-12600y+73305=1+16129#

or #625x^2+625y^2-4950x-12600y+73305-16130=0#

or #625x^2+625y^2-4950x-12600y+57175=0#

or #25x^2+25y^2-198x-504y+2287=0#

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Answer 2

Ella Armstrong

The equation of the circle is $(x - \frac{40}{13})^2 + (y - \frac{65}{13})^2 = \frac{260}{13}$

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.