Question

# The radius of the circle passing through the point $\left( 2,6 \right)$ two of whose diameters are $x+y=6$ and $x+2y=4$ is(A) $10$ (B) $2\sqrt{5}$ (C) $6$ (D) $4$

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Hint: The coordinates of the centre of the circle can be found out by finding the intersection points of the two diameters of the circle.

In the question, it is given $x+y=6$ and $x+2y=4$are diameters of the circle.
We know that diameters of the circle intersect only at the centre of the circle. So, the intersecting point of the above two diameters given in the question must be the centre. Hence, by solving the above to linear equations, we can find the centre of the circle.
Subtracting second equation and first equation;
\begin{align} & \left( x+2y \right)-\left( x+y \right)=4-6 \\ & \Rightarrow y=-2 \\ \end{align}
Substituting $y=-2$ in $x+y=6$, we get 🡪
\begin{align} & x-2=6 \\ & \Rightarrow x=8 \\ \end{align}
Hence, centre $\equiv \left( 8,-2 \right)$

Also, it is given in the question that the point $\left( 2,6 \right)$lies on the circle. We know that radius is equal to the distance between centre and any point on the circle.
To find the distance between the centre and the point, we will use the distance formula from which, the distance between any two coordinates $\left( x,y \right)$ and $\left( x',y' \right)$ is given by 🡪

$d=\sqrt{{{\left( x-x' \right)}^{2}}+{{\left( y-y' \right)}^{2}}}$

So, using the distance formula, the radius can be found by calculating distance between the centre of the circle (8,-2) and the point (2,6). Radius r is given by 🡪
\begin{align} & r=\sqrt{{{\left( 8-2 \right)}^{2}}+{{\left( -2-6 \right)}^{2}}} \\ & \Rightarrow r=\sqrt{{{\left( 6 \right)}^{2}}+{{\left( -8 \right)}^{2}}} \\ & \Rightarrow r=\sqrt{36+64} \\ & \Rightarrow r=\sqrt{100} \\ & \Rightarrow r=10 \\ \end{align}

Note: There is a possibility of mistake while calculating the distance by distance formula. There is a term in the calculation part. That term is ${{\left( -2-6 \right)}^{2}}$. There is a possibility that one by mistake writes it equal to ${{\left( -4 \right)}^{2}}$ instead of ${{\left( -8 \right)}^{2}}$.