Question

find the equation of the circle passing through (2,-1) having the centre at (2,3)?

Answer 1

Hi Mate!!

General equation of a circle = (x-h)² + (y-k)² = r²

Where ( h,k ) are the co-ordinates of center of the circle and r is radius.

h =2 and k = 3

r = √ {(2-2)²+(3+1)²}. by using distance formula

(x -2)² + (y-3)² = 4

Have a great future ahead...

Answer 2

The equation of the circle is

(x -2)² + (y-3)² = 16

Step 1: Given data  

The equation of the circle passing through (2,-1) having the center at ( 2,3).

Step 2: To find

The equation of the circle for the given data.

Step 3: Formula used

We know that,

Equation of a circle = ($x_{1}$-$x_{0}$)² + ($y_{1} -y_{0}$)² = r²

Where, ( $x_{0} ,y_{0}$ ) are the co-ordinates of center of the circle and r is radius.

To find the radius of the circle,

$r=\sqrt{(x_{0} -x_{1})^{2}+(y_{0} -y_{1} )^{2}$  

$r=\sqrt{(2-2_{})^{2}+(3-(-1)_{} )^{2}$

  r  =$\sqrt{4^{2} }$

r=4

Hence the equation of the circle is

(x -2)² + (y-3)² = 16

To learn more..

1) Find the equation of the circle passing through (-2,3) and having centre at (0,0)

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2) Find the equation of a circle whose centre is (2,-1) and which passes through the point (3,6).​

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