analytic geometrycirclesgeometry
Find the equations of the circles which are tangent to the x-axis, with radius of 5 units and passing through the point $(0,8)$.
I know how to formulate an equation with the given radius of 5 and then substitute the point $(0,8)$, but then I get stuck at finding their different centers.
The equation of your circle is then $(x-r)^2+(y-r)^2=r^2$. Plug in $x=2,y=1$ and solve for $r$.
Let us consider the centers $C$ and $D$ of these circles. Both circles have to go through $A,B$, so both $C$ and $D$ lie on the perpendicular bisector of $AB$. The distance of $C$ from $A$ equals the distance of $C$ from the $y$-axis. The distance of $D$ from $A$ equals the distance of $D$ from the $y$-axis, so both $C$ and $D$ lie on the parabola having the $y$-axis as directrix and the point $A$ as focus. It follows that the problem is solved by intersecting a line ($y=5-x$) and a parabola ($x=\frac{1+y^2}{2}$).
From $C(1,4)$ and $D(5,0)$ it follows that $r_C=1$ and $r_D=5$. These solutions can be found by educated guess, too: $C$ is trivially the center of a circle through $A$ and $B$ which is tangent to the $y$-axis. $D$ clearly works as a center since $3,4,5$ is a Pythagorean triple.
Alternative approach: let $P$ be the intersection of the $AB$-line with the $y$-axis. Let $Q,R$ be the tangency points of our circles with the $y$-axis. By considering the powers of $P$ with respect to our circles we have $$ PQ^2 = PR^2 = PA\cdot PB = \sqrt{2}\cdot 2\sqrt{2} = 4 $$ so $PQ=PR=2$, the locations of $P,Q$ are straightforward to be found and the locations of $C,D$ too.
Best Answer
The circle's centre will be $(k, 5)$ So the equation is $$(x-k)^2 + (y-5)^2 = 25.$$
Now put $(0,8)$ in the equation, to get $k = \pm 4$.