[Math] Find the equation of the circle which touches the $X$ axis at t

circles

Find the equation of the circle which touches the $X$ axis at the point $(4,0)$ and passes through the point $(2,2)$ .

My Attempt:

since the circle touches the $x$ axis, y- co ordinate of the centre is equal to the radius of circle. But, I don't see the centre of the circle given here. How could this be solved?

Best Answer

It's easiest to work with the equation of the circle given by

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

as this is the circle with center $(a,b)$ and radius $r$.

You are given two facts:

  1. The circle touches the $x$ axis at $(4,0)$.
  2. The circle passes through the point $(2,2)$.

The first fact tells you that the center of the circle must have an $x$ coordinate of $4$ and that the radius of the circle is the same as the $y$-coordinate.

Your job: Try to put the two things into equations. Knowing that $a$ is the $x$-coordinate and $b$ the $y$-coordinate of the center, and that $r$ is the radius, what equation does the sentence "The $x$ coordinate of the center is $4$" describe? What about the sentence "The radius is the same as the $y$ coordinate"?

The second fact tells you that plugging in $x=2$ and $y=2$ into the circle equation will result in a true statement. This should give you the third equation (the first fact gives you two), and you will have $3$ equations for $3$ variables, $a,b$ and $r$.

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