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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