Question

find the equation of the circle passing through the points (2, 3) and (-1, 1) and whose Centre is on the line x-3Y-11=0​

Answer 1

Step-by-step explanation:

Given C (h, k) be the centre and r the radius of circle. Let P(x, y) be any point on the circle. Then, by the definition, | CP | = r . By the distance formula, we have, (x−h)2+(y−k)2=r2

Now, given two pairs of co-ordinates through which the circle passes and the equation of the line through which the circle passes, we can substitute those values in the equation of the circle and the line and solve for h, k and r.

Given two pairs of co-ordinates through which the circle passes and the equation of the line through which the circle passes, we can substitute those values in the equation of the circle and the line and solve for h, k and r.

Given that the circle passes through (2,3)→(2−h)2+(3−k)2=r2...(1)

Given that the circle passes through (−1,1)→(−1−h)2+(1−k)2=r2...(2)

From (1) and (2), we get, ((2−h)2+(3−k)2=(−1−h)2+(1−k)2→6h+4k=11..(3)

Since the centre of the circle lines on x−3y−11=0→h−3k=11...(4)

Solving (3) and (4), we get, h=72 and k=−52

Substituting the values of h and k in (1), we get: (2−72)2+(y−−52)2=r2→−322+1122=r2

⇒94+1214=r2=652=r2

Now, we can substitute for h,k and r in (x−h)2+(y−k)2=r2 to get the equation of the circle:

(x−72)2+(y+52)2=652

4x2−28x+49+4y2+20y+25=130→4x2+4y2−28x+20y−56=0

x2+y2−7x+5y−14=0