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

\(\left( {6,- 6} \right), \left( {3, - 7} \right){\text{ and }} \left( {3,3} \right).\)

Text Solution

Reasoning:

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

Distance Formula 

\(\text=\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}\)

What is known?

The \(x\) and \(y\) co-ordinates of the points.

What is unknown?

The centre of the circle passing through the points

\(\left( {6,- 6} \right), \left( {3, - 7} \right){\text{ and }} \left( {3,3} \right).\)

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.

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

From the figure that

\(\begin{align}OA &= OB \\ ( \text{radii of}  & \text{ the same circle} ) \end{align}\)

\[\begin{align} \sqrt {\! (x\! - \!6)^2\! +\! (y \! +\! 6)^2 } & \! = \! \sqrt {\! (x \! -\! 3)^2 \!\! + \! (y \! + \! 7)^2 } \\\begin{bmatrix} {x^2} + 36 - 12x +\\ {y^2} + 36 + 12y \end{bmatrix} &\!=\! \begin{bmatrix} {x^2} + 9 - 6x +\\ {y^2} + 49 + 14y \end{bmatrix} \\\\\text{(Squaring on } & \text{both sides)}\\\\- 6x - 2y + 14 &= 0 \\ 3x + y &= 7 \ldots \text{Equation} (1) \end{align}\]

Similarly,

\(\begin{align}OA & = OC \\ ( \text{radii of} & \, \text{ the same circle} ) \end{align}\)

\[\begin{align} \! \sqrt { \! (x \! - \! 6)^2 \! + \! (y \! +\! 6)^2 } & \! = \! \sqrt {\! (x \! - \! 3)^2 \!+\! (y\! - \! 3)^2 }\\\begin{bmatrix} {x^2} + 36 - 12x + \\{y^2} + 36 + 12y \end{bmatrix} &= \begin{bmatrix} {x^2} + 9 - 6x + \\{y^2} + 9 - 6y \end{bmatrix}\\\\\text{(Squaring on } &\text{both sides)}\\\\- 6x + 18y + 54 &= 0 \\ - 3x + 9y &= - 27 \ldots(2) \end{align}\]

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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