# Ex.7.4 Q3 Coordinate Geometry Solution - NCERT Maths Class 10

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## Question

Find the centre of a circle passing through the points \begin{align}\left( {6,\, - {\text{ }}6} \right),{\text{ }}\left( {3, - {\text{ }}7} \right){\text{ and }}{\text{ }}\left( {3,\,\,3} \right).\end{align}

## Text Solution

Reasoning:

The distance between the two points can be measured using the Distance Formula which is given by:

Distance Formula       $$\text=\sqrt{\left(\mathrm{x}_{1}-\mathrm{x}_{2}\right)^{2}+\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)^{2}}$$

What is known?

The $$x$$ and $$y$$ co-ordinates of the points.

What is unknown?

The centre of the circle passing through the points \begin{align}\left( {6,\, - {\text{ }}6} \right),{\text{ }}\left( {3, - {\text{ }}7} \right){\text{ }}{\text{and }}\left( {3,\,\,3} \right).\end{align}

Steps:

From the Figure,

Given,

• Let \begin{align}O\left( {x,y} \right)\end{align} be the centre of the circle.
• Let the points \begin{align}\left( {6,\, - 6} \right),\,\,\left( {3, - 7} \right),\,\,\text{and}\left( {3,3} \right)\end{align} be representing the points A, B, and C on the circumference of the circle.

Distance from centre $$O$$ to $$A$$, $$B$$, $$C$$ are found below using the Distance formula mentioned in the Reasoning.

\begin{align}∴ {\text{OA}}& = \sqrt {{{({\text{x}} - 6)}^2} + {{({\text{y}} + 6)}^2}} \\{\text{ OB}} &= \sqrt {{{({\text{x}} - 3)}^2} + {{({\text{y}} + 7)}^2}} \\{\text{ OC}} &= \sqrt {{{({\text{x}} - 3)}^2} + {{({\text{y}} - 3)}^2}} \end{align}

From the figure that \begin{align}OA = OB \qquad \rm(radii \;of \;the \;same \;circle)\end{align}

\begin{align} \sqrt {{{(x - 6)}^2} + {{(y + 6)}^2}} &= \sqrt {{{(x - 3)}^2} + {{(y + 7)}^2}} \end{align}

\begin{align} {x^2} + 36 - 12x + {y^2} + 36 + 12y &= {x^2} + 9 - 6x + {y^2} + 49 + 14y\,\,\,\,\,\,\,\,\,\,\,\, \end{align} (Squaring on both sides)

\begin{align} - 6x - 2y + 14 &= 0 \end{align}

\begin{align}\qquad\qquad 3x + y &= 7 \qquad \qquad \qquad \ldots \ldots \end{align} Equation (1)

Similarly, \begin{align}OA = OC & \qquad \rm{}(radii\; of\; the \;same \;circle) \end{align}

\begin{align} \sqrt {{{(x - 6)}^2} + {{(y + 6)}^2}} &= \sqrt {{{(x - 3)}^2} + {{(y - 3)}^2}} \end{align}

\begin{align} {x^2} + 36 - 12x + {y^2} + 36 + 12y &= {x^2} + 9 - 6x + {y^2} + 9 - 6y\,\,\,\,\,\,\,\, \end{align} (Squaring on both sides)

\begin{align} - 6x + 18y + 54 &= 0 \end{align}

\begin{align}\quad - 3x + 9y &= - 27\quad \qquad \qquad \ldots \ldots \end{align} Equation(2)

On adding Equation (1) and Equation (2), we obtain

\begin{align}10y &= - 20\\y &= - 2\end{align}

From Equation (1),we obtain

\begin{align}3x - 2{\text{ }} = {\text{ }}7\\3x = {\text{ }}9\\x = {\text{ }}3\end{align}

Therefore, the centre of the circle is $$(3, -2)$$.

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