Question

If p=(-3,-2) and q=(1,6) are the endpoints of the diameter of a circle find the equation of the circle

Answer 1

Answer:

The equation of the circle   (x +1) )² +(y-(2))² = (2(√5))²

  or

The equation of the circle    x² + 2 x + y² - 4 y  = 15

Step-by-step explanation:

Given points end  Points are p(-3,-2) and q( 1,6)

The distance of two points formula

P Q = $\sqrt{x_{2} - x_{1})^{2} + ((y_{2} -y_{1})^{2} }$

P Q = $\sqrt{1 - (-3)^{2} + ((6 -(-2))^{2} }$

P Q = $\sqrt{16+64} = \sqrt{80}$

The diameter 'd' = 2 r

                       2 r = √80

                            = $\sqrt{16 X 5}$

                           = $4 \sqrt{5}$

                      r =  2√5

Mid-point of two end points

                       $(\frac{x_{1} + x_{2} }{2} , \frac{y_{1} +y_{2} }{2} ) = (\frac{-3+1}{2} ,\frac{-2 +6}{2} )$

                                              = (-1 ,2)

Mid-point of two end points = center of the circle

                                    (h,k) = (-1 , 2)

                 

The equation of the circle

           (x -h )² +(y-k)² = r²

         (x -(-1) )² +(y-(2))² = (2(√5))²

        x² + 2 x + 1 + y² - 4 y + 4 = 20

         x² + 2 x + y² - 4 y  = 20 -5

          x² + 2 x + y² - 4 y  = 15

Final answer:-

The equation of the circle   (x +1) )² +(y-(2))² = (2(√5))²

  or

The equation of the circle    x² + 2 x + y² - 4 y  = 15