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Question

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A. ${x^2} + {y^2} = 1$

B. ${x^2} + {y^2} - 2x + 1 = 0$

C. ${x^2} + {y^2} - 2x - 1 = 0$

D. ${x^2} + {y^2} = 5$

Answer
Verified

Let us note down the given data,

A foot of normal from the point $\left( {4,3} \right)$ to a circle is $\left( {2,1} \right)$

Diameter of the circle has an equation $2x - y = 2$.

Let us draw a diagram to visualize it clearly,

Now we can notice that the points $\left( {4,3} \right)$ and $\left( {2,1} \right)$ lie on the normal, we can find out the equation of the normal.

So, the equation of the normal can be found through the above formula.

Slope of the normal $ = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$

Substitute the values we get,

Slope of the normal $ = \dfrac{{1 - 3}}{{2 - 4}}$

$ = 1$

The equation of a line which passes through two points $\left( {{x_1},{y_1}} \right)$ and $\left( {{x_2},{y_2}} \right)$ with slope $m$ is \[\left( {y - {y_1}} \right) = m\left( {x - {x_1}} \right)\].

So, the equation of the normal is $\left( {y - {y_1}} \right) = m\left( {x - {x_1}} \right)$

When we substitute the values, we get as follows:

The equation of normal,

$

\Rightarrow y - 1 = 1\left( {x - 2} \right) \\

\Rightarrow y - 1 = x - 2 \\

\Rightarrow x - y - 1 = 0 \\

$

Now, to get the centre, we need to solve the normal equation and the diameter equation.

In the normal equation,

$

\Rightarrow x - y - 1 = 0 \\

\Rightarrow x = 1 + y \\

$

Put this in the diameter equation.

Diameter of the circle is $2x - y = 2$

$

\Rightarrow 2\left( {1 + y} \right) - y = 2 \\

\Rightarrow 2 + y = 2 \\

\Rightarrow y = 0 \\

$

Put this value in the normal equation or diameter equation to get the centre coordinators. Here we are going to use the normal equation. You can use any equation. Both give the same answer.

$

\Rightarrow x = 1 + y \\

\Rightarrow x = 1 + 0 \\

\Rightarrow x = 1 \\

$

So, the coordinates of the centre of the circle are $\left( {1,0} \right)$.

Now the equation of the circle with centre $\left( {a,b} \right)$ and radius $r$ is \[{\left( {x - a} \right)^2} + {\left( {y - b} \right)^2} = {r^2}\]

So, the circle equation will be as follows:

\[

\Rightarrow {\left( {x - 1} \right)^2} + {\left( {y - 0} \right)^2} = {r^2} \\

\Rightarrow {\left( {x - 1} \right)^2} + {y^2} = {r^2} \\

\]

We know that the point $\left( {2,1} \right)$ is on the circle. So, substitute this point in the equation to get the value of radius.

\[

\Rightarrow {\left( {2 - 1} \right)^2} + {1^2} = {r^2} \\

\Rightarrow 1 + 1 = {r^2} \\

\Rightarrow {r^2} = 2 \\

\Rightarrow r = \sqrt 2 \\

\]

Substitute this value in the equation of circle, we will get \[{\left( {x - 1} \right)^2} + {y^2} = 2\]

Expand the equation using the formula ${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$.

Where $a = x$ and $b = 1$.

Then we will get,

\[

{\left( {x - 1} \right)^2} + {y^2} = 2 \\

\Rightarrow {x^2} + 1 - 2x + {y^2} = 2 \\

\Rightarrow {x^2} + {y^2} - 2x - 1 = 0 \\

\]

So, the equation of the circle is \[{x^2} + {y^2} - 2x - 1 = 0\]