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Question

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A.${a^2}$

B.${b^2}$

C.${r^2}$

D.None of these

Answer

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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 \[\lambda = {r^2}\]

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