Math, asked by ektavarshney878, 7 months ago

find the center and radius of circle represented by xsquare +ysquare+6x-4y+4=0​

Answers

Answered by pulakmath007
1

SOLUTION :-

TO DETERMINE :-

The center and radius of the circle given by

 \sf{ {x}^{2}  +  {y}^{2} + 6x - 4y + 4 = 0 }

FORMULA TO BE IMPLEMENTED :-

For the circle of the form

 \sf{ {(x - a)}^{2}  +  {(y - b)}^{2}  =  {r}^{2} }

The center of the circle is ( a, b ) and

radius of the circle is r unit

EVALUATION :-

Here the given equation of the circle is

 \sf{ {x}^{2}  +  {y}^{2} + 6x - 4y + 4 = 0 }

Which can be rewritten as below

 \sf{ {x}^{2}  +  {y}^{2} + 6x - 4y + 4 = 0 }

 \implies \sf{ {x}^{2} + 6x  +  {y}^{2}  - 4y + 4 = 0 }

 \implies \sf{ {x}^{2} + 6x   + 9+  {y}^{2}  - 4y + 4 = 9 }

 \implies \sf{ {(x)}^{2} + 2.x.3+  {(3)}^{2} +  {y}^{2}  - 2.y.2 +  {(2)}^{2}  =  {(3)}^{2}  }

 \implies \sf{ {(x + 3)}^{2} +  {(y - 2)}^{2} =  {(3)}^{2}  }

Which is of the form

 \sf{ {(x - a)}^{2}  +  {(y - b)}^{2}  =  {r}^{2} }

Hence the required center of the circle is

( - 3 , 2 ) and radius of the circle is 3 unit

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