Get Answers to all your Questions

header-bg qa

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

Answers (1)

best_answer

As we know, 

The equation of the circle with centre ( h, k) and radius r is given by ;

(x-h)^2+(y-k)^2=r^2

Given Here,

Condition 1: the circle passes through points   (2,3) and (–1,1)

(2-h)^2+(3-k)^2=r^2

(-1-h)^2+(1-k)^2=r^2

Here, 

(2-h)^2+(3-k)^2=(-1-h)^2+(1-k)^2

(2-h)^2-(-1-h)^2+(3-k)^2-(1-k)^2=0

(3)(1-2h)+(2)(4-2k)=0

3-6h+8-4k=0

6h+4k=11

Now, Condition 2: centre is on the line.x - 3y - 11 = 0

h-3k=11

From condition 1 and condition 2 

h=\frac{7}{2},\:k=\frac{-5}{2}

Now let's substitute this value of h and k in condition 1 to find out r

\left ( 2-\frac{7}{2}\right )^2+\left (3+\frac{5}{2}\right )^2=r^2

\frac{9}{4}+\frac{121}{4}=r^2

r^2=\frac{130}{4}

So now, the Final Equation of the circle is 

\left(x-\frac{7}{2}\right )^2+\left(y+\frac{5}{2}\right)^2=\frac{130}{4}

x^2-7x+\frac{49}{4}+y^2+5y+\frac{25}{4}=\frac{130}{4}

x^2+y^2-7x+5y-\frac{56}{4}=0

x^2+y^2-7x+5y-14=0

Posted by

Pankaj Sanodiya

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads