Equation of circles tangent to $x$-axis with a given radius and point

Question

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.

Answer 1

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 $k = \pm 4$.

Answer 2

HINT:

Let the equation of the circle be $$(x-a)^2+(y-b)^2=5^2$$

Let us find the intersection with $x$ axis i..e, $y=0$

$$(x-a)^2+(0-b)^2=5^2\iff x-a=\pm\sqrt{5^2-b^2}$$

As the roots of the above equation represent the abscissa of the intersection,

for tangency, we need the roots to be same $\implies25-b^2=0$

Now use the fact that the circle passes through $(0,8)$

Answer 3

The equation of circle is $(x-a)^2+(y-b)^2=r^2$ centered ar point $(a,b)$.

The equation of line is $y=mx+c$. The x axis is the line $y=0$.

All circles having tangent to x axis implies it have point $y=0$ and only one solution for x.

$(x-a)^2+b^2=5^2$

$x^2-2ax+a^2+b^2-25=0$

We want solutions for x. We thus require the discriminant to be zero. $b^2-4ac$ is $26-a^2-b^2=0$.

These are the equation which is to be solved for to find $a,b$ for single solutions. $b^2=26-a^2$

Inserted into equation of circle with $y=0$

$x^2-2ax+a^2+26-a^2-25=0$

$x^2-2ax+1=0$

Answer 4

Use the tangent squared = segments product property of circle from an external point to determine coordinates of points on y-axis as

$$ ( 0,8),(0,h^2/8)$$

with tangent point on x-axis

$$ (h,0)$$

Distance squared between $y-$ axis point and circle center

$$ ( (h-0)^2 +(8-5)^2 = 5^2 \rightarrow h=\pm 4 $$

The two solution circles have equations:

$$ ( x \pm 2)^2 +y^2 = 25 $$

The two circles also pass through the point $(0,2)$ on $y-$ axis.

Answer 5

Use the tangent squared = segments product property of circle from an external point to determine coordinates of points on y-axis as

$$ ( 0,8),(0,h^2/8)$$

with tangent point on x-axis

$$ (h,0)$$

Distance squared

$$ ( (h-0)^2 +(8-5)^2 = 5^2 \rightarrow h=\pm 4 $$

The two solution circles have the equations:

$$ ( x \pm 2)^2 +(y-5)^2 = 25. $$

$$ ( x \pm 2)^2 + y^2 -10 y =0 $$

They also pass through the point $(0,2)$ on $y-$ axis.