Find the circle's equation that passes through A(10,14),
The circle tangents to the parabola $$y^2= 16x$$ at P(16,16).
I have tried to find the center point of the circle:
Comparing the radiuses: OA and OP.
By this way, b = 54 - 3a.
so the center point of the circle is: (a, 54-3a).
There is a sentence:
The equation of a tanget to a circle: $$(x-a)^2 + (y-b)^2 = R^2$$ at the point of $$(x_0,y_0)$$ is:
$$(x_0-a)(x-a)+(y_0-b)(y-b)=R^2$$
That's why:
$$(16-a)(x-a)+(16-[54-3a])(y-[54-3a])=R^2$$
I don't know how to continue..
The answer to this exercise is:
$$(x-6)^2 + (y-36)^2 = 500$$
Note that (6,36) solves the equation: b = 54 - 3a, so this equation is right.
The tangent line to a circle at some point on the circle is always perpendicular to a radius to that point. This means that its center lies somewhere on a line perpendicular to the tangent line. By implicitly differentiating the equation of the parabola, we can find that the slope of the tangent to its positive branch at a point $P(x,y)$ is $m=\frac8y$. The slope of a line perpendicular to this is then $-\frac1m=-\frac y8$ and so the center of the circle lies somewhere on the line $(y-16)=-2(x-16)$, or $y=-2x+48$.
As you know, the equation of a circle centered at $C(x_c,y_c)$ with radius $r$ is $$(x-x_c)^2+(y-y_c)^2=r^2.$$ If we substitute the two known points and the equation of the line on which the center lies into this general equation and then equate the two left sides, we get $$(16-x_c)^2+(16-(48-2x_c))^2=(10-x_c)^2+(14-(48-2x_c))^2$$ which gives $x_c=6$. Computing $y_c$ and $r$ is now a straightforward calculation.
A slightly different approach is to write the equation of the center line in vector form as $\vec v=t\vec n+P$, where $\vec n=\left(\frac1{\sqrt5},-\frac2{\sqrt5}\right)$, a unit vector that’s perpendicular to the parabola at $P$. This vector can be computed from the slope of this line. The advantage of normalizing the direction vector is that the parameter $t$ then represents the distance from the point $P$, so when $\vec v$ is the center of our circle $t$’s absolute value will be equal to the circle’s radius. So, we now have the parametric equations $$\begin{align}x&=\frac 1{\sqrt5}t+16\\y&=-\frac 2{\sqrt5}t+16\end{align}$$ for the line on which the center lies. Substituting these expressions and the known point $A$ into the general equation of a circle gives the equation $$\left(10-\left(\frac1{\sqrt5}t+16\right)\right)^2+\left(14-\left(16-\frac2{\sqrt5}t\right)\right)^2=t^2$$ which has the solution $t=-10\sqrt5$. From here, computing the radius and center of the circle, and so finding its equation, is again a straightforward computation.