Question

The center of a circle lies on x+y=3 and touching the lines x=3, y=3 then find diameter of circle ?

Answer 1

Let O(x, y) be center of the circle.

The line $x=3$ is parallel to y axis. If we draw a line parallel to x axis from O rightwards, it meets the line $x=3$ at A(3, y). This point lies on circle so OA is a radius of the circle.

Here x < 3.

The line $y=3$ is parallel to x axis. If we draw a line parallel to y axis from O upwards, it meets the line $y=3$ at B(x, 3). This point lies on circle so OB is radius of the circle.

Here y < 3.

Thus,

$\longrightarrow OA=OB$

$\longrightarrow\sqrt{(3-x)^2+(y-y)^2}=\sqrt{(x-x)^2+(3-y)^2}$

$\longrightarrow\sqrt{(3-x)^2}=\sqrt{(3-y)^2}$

$\longrightarrow3-x=3-y$

$\longrightarrow x=y\quad\quad\dots(1)$

Since the center lies in $x+y=3,$

$\longrightarrow x+y=3$

From (1),

$\longrightarrow2x=2y=3$

$\longrightarrow x=y=1.5$

Hence diameter of the circle is,

$\longrightarrow d=2(3-x)$

$\longrightarrow\underline{\underline{d=3}}$