Question

Question 11 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.

Class X1 - Maths -Conic Sections Page 241

Answer 1

hello users...
solution:-
we know that:
the equation of circle is
 x² + y²  +2gx +2fy + c = 0  ...(1.)

where;
(-g,-f) is centre. 

=> -g - 3(-f) - 11 = 0  
=> 3f - g - 11 = 0
( because the given equation is x – 3y – 11 = 0 , and (x,y) as centre )

=> g = 3f - 11  ....(2.)

As
point (2,3) lies on the equation of circle (1.)

=> 2² + 3² + 2×2 ×g + 2× 3×f  + c = 0

=> 13 + 4g + 6f + c = 0 

=> 13 + 4(3f - 11) + 6f + c + 0  ......from (2.)

=> 13 + 12f - 44 + 6f + c = 0

=>  -31 + 18f + c = 0

=> c = 31 - 18f   ...(3)

Also,
point (-1,1) lies on equation of circle (1.)

=> (-1)² + 1² + 2×(-1) g + 2 ×1 f + c = 0

=> 1+1 -2g +2f + c = 0

=> 2 -2 (3f - 11) + 2f + c = 0   ....from (2.)

=> 2 - 6f +22 +2f +c = 0 

=> 24 -4f +c = 0 

=> c = 4f - 24 ...(4.)

Now,
comparing (3.) and (4.) we get
31 - 18f = 4f - 24

=> 18f + 4f = 31 + 24 

=> 22f = 55

=> f = 5 / 2

now,
 putting the value of f in equation (3.) , we get 
=> c = 31 - 18 × 5/2 

=> c = 31 - 45 

=> c = -14

and also
putting the value of f in equation (2.) we get
=> g = 3 * 5/2  - 11  

=> g = -7 / 2

Now
putting the value of g , f and c in equation (1.) we get
=> x² + y² + 2×(-7/2) x + 2× 5/2 y + (-14) = 0

=> x² + y² - 7x + 5y -14 = 0 Answer

@ hope it helps :)

Answer 2

Let equation of circle is
x² + y² + 2gx + 2fy + C = 0 , then centre of circle is ( -g, - f) .

A/C to question ,
(2, 3) and (-1, 1) passing through the circle. e.g they will satisfy it .

when (2, 3)
(2)² + (3)² + 2g(2) + 2f(3) + C = 0
4 + 9 + 4g + 6f + C = 0
4g + 6f + C + 13 = 0 __________(1)


when, (-1,1)
(-1)² + (1)² + 2g(-1) + 2f(1) + C = 0
1 + 1 -2g + 2f + C = 0
-2g + 2f + C + 2 = 0____________(2)

again, question said centre of circle is on the line x - 3y - 11 = 0 . so, centre ( -g, -f) will satisfy equation of line .

so, (-g) -3(-f) -11 = 0
-g + 3f - 11 = 0___________(3)


now,
subtract equations (2) from (1)
4g + 6f + C + 13 - (-2g + 2f + c + 2 ) = 0
6g + 4f + 11 = 0__________(4)

now, solve equations (3) and (4)
we get f = 5/2 and g = -7/2 , put it in equation (1)
4 × (-7/2) + 6 × (5/2) + C + 13 = 0
-14 + 15 + C + 13 = 0
C = -14

now, put values of g, f and C in equation of circle.

x² + y² + 2x(-7/2) + 2y(5/2) + (-14) = 0
x² + y² -7x + 5y -14 = 0

hence, equation of circle is
x² + y² -7x + 5y -14 = 0