Question

What’s the equation of the circle passing through the points (0, 0), (0, 1) (6, 0)

• A. $x^2+y^2+6x+6y=0$
• B. $x^2+y^2-3x+12y=0$
• C. $x^2+y^2-6x-y=0$
• D. $x^2+y^2+x+y=0$

Answer 1

The correct option is C

$x^2+y^2-6x-y=0$

What is the first method that comes to your mind for finding the equation of a circle that passes though 3 points.? We used to take the general equation of the circle, then we put the points into that equation to find the constants and then arrive at the final required equation of the circle. But instead of this multi step process we will be able to reach that equation using a simple step using determinants. For that we use the below formula.

$\left|\begin{array}{cccc}{x}^2+y^2& x& y& 1\\ x_1^2+y_1^2& x_1& y_1& 1\\ x_2^2+y_2^2& x_2& y_2& 1\\ x_3^2+y_3^2& x_3& y_3& 1\end{array}\right|=0$

In the formula we put the points to obtain the expression below,

$D=\left|\begin{array}{cccc}{x}^2+y^2& x& y& 1\\ 0& 0& 0& 1\\ 1& 0& 1& 1\\ 36& 6& 0& 1\end{array}\right|=0$

Points are

A = (0, 0)

B ≡ (0, 1)

C ≡ (6, 0)

$D=1\left|\begin{array}{ccc}{x}^2+y^2& x& y\\ 1& 0& 1\\ 36& 6& 0\end{array}\right|=0$

$ie.,(x^2+y^2)(-6)-x(-36)+y\left(6\right)=0$
$x^2+y^2-6x-y=0$ .This is the required equation of the circle.