Question

Prove that if (x, y) be a point on the circle with the line joining (0, 1) and (2, 3) as diameter, then x² + y² – 2x – 4y + 3 = 0. Find the coordinates of the points where this circle cuts they axis.

Answer 1

As (0, 1), (2, 3) are the points joining the diameter of the circle, the center will be

\(=(\frac{0+2}{2},\frac{1+3}{2})=(1,2)\)

Radius = \(\sqrt{(1-0)^2+(2-1)^2}\)

\(\Rightarrow\) r = \(\sqrt{1^2+1^2}=\sqrt{2}\)

(x, y) lies on the cirlce

(x - 1)² + (y - 2)² = \(\sqrt{2}^2\)

x² - 2x + 1 + y² - 4y + 4 = 2

x² + y² - 2x - 4y + 5 - 2 = 0

x² + y² - 2x - 4y + 3 = 0

Let (x, 0) be the coordinate which intersect the x axis then, x² – 2x + 3 = 0

\(x=\frac{2\pm\sqrt{4-12}}{2}=\frac{2\pm\sqrt{-8}}{2}\)

This circle can not be intersect the x axis. 

Let (0, y) be the coordinate which intersect the y axis then, y – 4y + 3 = 0, (y – 3)(y – 1) = 0 

Coordinates are (0, 1) and (0, 3)