Question

If (-1,4) and (3,-2) are end points of diameter of the circle, then the equation of the circle is

Answer 1

Answer:

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Step-by-step explanation:

If the centre of a circle is (a, b) and radius r, then the equation of the circle is :

$(x - x)^2 + (y-b)^2 = r^2$

First we will find the centre, (a , b):

$a = \frac{3+(-1)}{2} = \frac{2}{2} = 1\\\\b = \frac{-2 + 4}{2} = \frac{2}{2} = 1$

Find radius:

$Radius, r = \frac{Diameter}{2}$

Diameter is the distance between (-1, 4) and (3, -2)

$diameter = \sqrt{(x_2 - x_1)^2 +(y_2-y_1)^2}$

             $= \sqrt{(3-(-1)^2 + (-2-4)^2} \\\\=\sqrt{16 + 36}\\\\=\sqrt{52}\\\\=\sqrt{4 \times 13}\\\\=2\sqrt{13} \ units$

Therefore , $r = \frac{2 \times \sqrt{13} }{2} = \sqrt{13} \ units$

Equation of the circle is :  

                                 $(x - 1)^2 +(y-1)^2 = 13$