Question

Find the equation of circle passing through (7,3) having radius 3 and whose center is on the line y=x-1

Answer 1

hi dear,
here is ur answer..................................
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general equation of circle:x^2+y^2+2gx+2fy+c=0
it passes through point (7,3) so this point must satisfy this equation...
after putting values that is x=7 and y=3 we get
14g+6y+c=-58.............................1.
as we know the centre of circle is (-g,-f) so this must satisfy the given equation of line on which centre lies
f=g+1...........................................2.
radius^2=g^2+f^2-c
that is....
9=g^2+f^2-c
put the values from equation 1 and 2
we will get  a quadratic equation in form
g^2+11g+28=0
so after this the value of g=-4 or-7
f=g+1
so, the values of f=-3 or -6
and c=-58-14g-6f
   c =16 or 76
put these values in the general equation of the circle
so,
equations of circle are........
x^2+y^2-8x-6y+16=0
         or
x^2+y^2-14x-12y+76=0

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HOPE IT HELPS UH..............
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Answer 2

answer is 0 ................