Question

Find the equation of the circle passing through the points (2, 3) and (–1, 1) and whose centre is on the line x – 3 y – 11 = 0.

Answer 1

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