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.
$\endgroup$ 15 Answers
$\begingroup$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 $k = \pm 4$.
$\endgroup$ 2 $\begingroup$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 $(0,8)$
$\endgroup$ $\begingroup$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 $y=0$
$x^2-2ax+a^2+26-a^2-25=0$
$x^2-2ax+1=0$
$\endgroup$ $\begingroup$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.
$\endgroup$ $\begingroup$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.
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