Question

If one end point of diameter of a circle
is (-2, 3) and centre of circle is (0,3)
then find the other end point of the
diameter​

Answer 1

The other end point of the  diameter​ is ( 2 , 3 ) .

Given :

One end point of diameter of a circle is ( -2 , 3 ) .

Center of circle ( 0 , 3 ) .

Let , the other end point of the diameter is ( x , y ) .

We know mid point between two points $(x_1,y_1)$  and $(x_2.y_2)$ is :

$X=\dfrac{x_1+x_2}{2}$

$Y=\dfrac{y_1+y_2}{2}$

Now ,

$0=\dfrac{x-2}{2}\\\\x=2\\\\3=\dfrac{y+3}{2}\\\\y=3$

The other end point of the  diameter​ is ( 2 , 3 ) .

Hence , this is the required solution .

Learn More :

Circle

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Answer 2

$Let \: A\: and \: B \: are \: two \:end \: points \\of \: diameter$

$O \:is\: the \: centre \: of \:circle .$

$Let \: A = ( x_{1}, y_{1}) , \\B( x_{2}, y_{2})= (-2,3) \: and \: O = ( 0,3)$

$\underline {\pink { Midpoint \: of \:AB segment : }}$

$Midpoint \: of\: diameter \:AB = O$

$\implies \big( \frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\big) = ( 0 , 3 )$

$\implies \big( \frac{ x -2 }{2}, \frac{y+3}{2}\big) = ( 0 , 3 )$

$\implies \frac{x-2}{2} = 0 , \: \frac{y+3}{2} = 3$

$\implies x-2= 0 , \: y+3= 6$

$\implies x = 2, \: y= 6 - 3$

$\implies x = 2, \: y= 3$

Therefore.,

$\red { Coordinates \: other \:end \:point }$

$\green { = ( 2 , 3 ) }$

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