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Answer:
Solution
The given points through which circle passes are ( 2,3 ) and ( −1,1 ) and the line on which center lies is x−3y=11 .
The equation of the required circle is given by,
( x−h ) 2 + ( y−k ) 2 = r 2 (1)
Where h and k denotes the center of the circle and r denotes the radius of the circle.
Since circle passes through the point ( 2,3 ) , this point will satisfy the equation of the circle represented in equation ( 1 ) .
( 2−h ) 2 + ( 3−k ) 2 = r 2 4+ h 2 −4h+9+ k 2 −6k= r 2 (2)
Since circle also passes through the point ( −1,1 ) , this point will also satisfy the equation of the circle represented in equation ( 1 )
( −1−h ) 2 + ( 1−k ) 2 = r 2 1+ h 2 −2h+1+ k 2 −2k= r 2 (3)
Equate equation ( 2 ) and ( 3 ) to determine the relation between h and k ..
4+ h 2 −4h+9+ k 2 −6k=1+ h 2 −2h+1+ k 2 −2k 4−4h+9−6k=1−2h+1−2k 6h+4k=11 (4)
Now since center lies on x−3y=11 , Center ( h,k ) will satisfy this equation.
So, h−3k=11 (5)
Solve equation ( 4 ) and ( 5 ) to determine the value of h and k ,
Multiply equation ( 5 ) by 6 and subtract from equation ( 4 ) we get,
6h+4k−11−( 6h−18k−66 )=0 6h+4k−11−6h+18k+66=0
Further simplify the equations,
22k=−55 k= −55 22 k= −5 2
Substitute the value of k in equation ( 5 ) ,
h+3( 5 2 )=11 h=11− 5 2 h= 7 2
Substitute the value of h and k in equation (1) to obtain the value of r 2 .
( 2− 7 2 ) 2 + ( 3+ 5 2 ) 2 = r 2 ( 4−7 2 ) 2 + ( 6+5 2 ) 2 = r 2 ( −3 2 ) 2 + ( 11 2 ) 2 = r 2
Further simplify the equations,
( 9 4 )+( 121 4 )= r 2 ( 130 4 )= r 2
Substitute the value of h and k and r 2 in equation (1) to obtain the equation of circle.
( x− 7 2 ) 2 + ( y+ 5 2 ) 2 = 130 4 ( 2x−7 2 ) 2 + ( 2y+5 2 ) 2 = 130 4 4 x 2 −28x+49+4 y 2 +20y+4 y 2 +25=130 4 x 2 −28x+4 y 2 +20y+4 y 2 −56=0
Further simplify the equations,
4( x 2 −7x+ y 2 +5y+ y 2 −14 )=0 x 2 −7x+ y 2 +5y+ y 2 −14=0
Thus the equation of the circle passing through ( 2,3 ) and ( −1,1 ) and whose center is on the line x−3y−11=0 is x 2 −7x+ y 2 +5y+ y 2 −14=0