Question

Find the equation to the circle whose center is at the point $\left( {\alpha ,\beta } \right)$and which passes through the origin. Also, prove that the equation to the tangent at the origin is $\alpha x + \beta y = 0$.

Answer 1

Hint: In order to solve the question, first using the general form of the equation of a circle find the required equation of the circle. Next, write the standard equation of a circle and write the equation of the tangent at point $\left( {{x_1},{y_1}} \right)$. Finally, prove that the equation to the tangent at the origin is $\alpha x + \beta y = 0$ using the above standard equation.

Formula Used:
The general form of the equation of a circle is
${\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}$
The standard equation of a circle is given by
${x^2} + {y^2} + 2gx + 2fy + c = 0$

Complete step by step solution:
Given that the center is at $\left( {\alpha ,\beta } \right)$.
We know the general form of the equation of a circle is
${\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}$
That is, we can write that
${\left( {x - \alpha } \right)^2} + {\left( {y - \beta } \right)^2} = {r^2}$. . . . . (1)
Since the circle passes through the origin we can write that
${\alpha ^2} + {\beta ^2} = {r^2}$. . . . . (2)
Substituting equation (2) in (1), that is
${\left( {x - \alpha } \right)^2} + {\left( {y - \beta } \right)^2} = {\alpha ^2} + {\beta ^2}$
We get
${x^2} - 2\alpha x + {\alpha ^2} + {y^2} - 2y\beta + {\beta ^2} = {\alpha ^2} + {\beta ^2}$
${x^2} + {y^2} - 2\alpha x - 2y\beta = 0$
This is the required equation of the circle.
Here it is also asked to prove that the equation to the tangent at the origin is $\alpha x + \beta y = 0$.
We know that the standard equation of a circle is given by
${x^2} + {y^2} + 2gx + 2fy + c = 0\;$
At point $\left( {{x_1},{y_1}} \right)$ the equation of the tangent is
$\;x{x_1} + y{y_1} + g(x + {x_1}) + f(y + {y_1}) + c = 0$
Here the point is at (0,0), so the equation of the tangent is
$gx + fy + c = 0$
That is
$g = - \alpha ,{\rm{ }}f = - \beta ,{\rm{ }}c = 0$
Since the center is at the point $\left( {\alpha ,\beta } \right)$.
That is
$\alpha x + \beta y = 0$

Hence the required equation is ${x^2} + {y^2} - 2\alpha x - 2y\beta = 0$ and also proved that the equation to the tangent at the origin is $\alpha x + \beta y = 0$.

Note: Students can get confused with the general and standard equation of a circle. Always remember that the center of a circle is the fixed point in the middle of the circle. Also, remember that when the center of the circle is at origin then the equation will be ${x^2} + {y^2} = 1$.