Question

The equation of the tangent to the circle \[{x^2} + {y^2} = {r^2}\]at $\left( {a,b} \right)$ is $ax + by - \lambda = 0$, where $\lambda $ is
A.${a^2}$
B.${b^2}$
C.${r^2}$
D.None of these

Answer 1

Hint: Here, we are required to find the value of $\lambda $, where it is given that a tangent touches the circle at point $\left( {a,b} \right)$ and the equations of both the tangent as well as the circle are given. Since, the point of contact is the intersection of both the tangent as well as the circle, it satisfies both the equations. Hence, we will substitute $\left( {x,y} \right) = \left( {a,b} \right)$ in both the equations of the circle as well as the tangent, and then, equating both of them, we will find the required value of $\lambda $.

Complete step-by-step answer:
Equation of circle: \[{\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}\], where $r$ is the radius of the circle and $\left( {h,k} \right)$ represents the centre of the circle.
The general equation of a circle is in the form of \[{\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}\] , where $r$ is the radius of the circle and $\left( {h,k} \right)$ represents the centre of the circle.
According to the question,
Equation of the circle is \[{x^2} + {y^2} = {r^2}\]
Hence, this can be written in the form of: \[{\left( {x - 0} \right)^2} + {\left( {y - 0} \right)^2} = {r^2}\]
This shows that $\left( {h,k} \right) = \left( {0,0} \right)$
Hence, the centre of the circle lies on the origin.
Now, when a tangent touches the circumference of a circle at point $\left( {a,b} \right)$,
Then, this point should satisfy the equation of the circle as well as the equation of the tangent.
This is because of the fact that this point acts as an intersection between the tangent and the circle, hence, satisfying the equation of both of them.
Therefore, since the equation of the circle is \[{x^2} + {y^2} = {r^2}\] and the tangent passes through it at the point $\left( {a,b} \right)$, hence, substituting $\left( {x,y} \right) = \left( {a,b} \right)$ in the equation of circle, we get,
\[{a^2} + {b^2} = {r^2}\]…………………………$\left( 1 \right)$
Now, according to the question,
Equation of the tangent is $ax + by - \lambda = 0$
Since, the point $\left( {a,b} \right)$ acts as an intersection, it must satisfy the equation of the tangent as well.
Therefore, substituting $\left( {x,y} \right) = \left( {a,b} \right)$ in the equation of the tangent $ax + by - \lambda = 0$, we get,
$a \cdot a + b \cdot b - \lambda = 0$
$ \Rightarrow {a^2} + {b^2} = \lambda $……………………$\left( 2 \right)$
Now, equating the equations $\left( 1 \right)$ and $\left( 2 \right)$, we get,
\[\lambda = {r^2}\]
Therefore, the required value of \[\lambda = {r^2}\]
Hence, option C is the correct answer.

Note: A circle is a shape drawn on a plane which consists of all the points which are equidistant from the given centre. The distance between each point on the circumference of the circle and the centre is called its radius. In a standard form, the equation of a circle is always written in the form of \[{\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}\] , where $r$ is the radius of the circle and $\left( {h,k} \right)$ represents the centre of the circle. Also, a tangent drawn to a circle is defined as a straight line which touches the circle at a single point on its circumference. The point where the tangent touches the circle is called the point of contact.