Question

find the eq of the circle which passes through the point (1,3)and (2,-1) and has its centre on the line 2x +y-4 =0

Answer 1

hello users ...

solution:-
let the equation of circle is

= x² + y² + 2gx + 2fy + c = 0  ....(1.)

where,
(-g , - f) is the centre 

=> -2g - f - 4 = 0

....( given equation is 2x + y -4 = 0  , and point (x,y) as centre)

=> f = -2g -4 ..(2.)

As,
(1,3) lies on equation of circle  (1.) 
=>1² + 3² +2×1×g + 2×3×f +c = 0

=> 1+9 +2g +6f +c = 0

=> 10 + 2g + 6f +c = 0  ....(3.)

=> 10 + 2g + 6(-2g -4) +c = 0  ..from equ. (2.)
=> 10 + 2g -12g - 24 +c = 0 
=> -14 -10g +c = 0   ...(4.)

=> c = 10g +14  ...(5.)
and,
As,
(2,-1) lies on equation of circle (1.)
=> 2² +(-1)² + 2×2×g + 2×(-1)×f + c = 0

=> 4+1 +4g - 2f +c = 0 

=> 5 +4g - 2f +c = 0   
=>5 + 4g -2(-2g -4) +c = 0   ...from equation (2.)
=> 5 + 4g +4g +8 +c = 0
=> 13 + 8g +c = 0 .....(6.)

=>c = -13 - 8g ..(7.)
now,
comparing  (5.) and (7.) 
we get.

=> 10g +14= -13 - 8g 

=> 10g + 8g = -13 - 14

=> 18g = -27 

=> g = -27/18 

Now,
putting the value of g in (2.) 
we get
=> f = -2×(-27/18) - 4 

=> f = 27/9 - 4 

=> f = 3 - 4   = -1

Now,
putting the value of g in equation (5.) 
we get,
=> c = 10×(-27/18)  + 14

=> c =  5×(-3) + 14
=> c = -15 +14  = -1

Now,
putting value of f,g and c in equation (1.)
We get,
=> x² + y² + 2×(-27 / 18) × x + 2×(-1) × y  + (-1) = 0 

=> x² + y² - 27 x / 9 - 2y   -1 = 0 

=> 9x² + 9y² - 27 x  -18y   - 9 = 0 Answer 

✮✭ hope it helps :) ✮✭