[ASK] About Circles in Coordinates

[Monoxdifly]

3. A(a,b), B(-a,-b), and C is plane XOY. P moves along with curve C. If the multiplication product of PA's and PB's gradients are always k, C is a circle only if k = ...?
4. The radius of a circle which meets X-axis at (6,0) and meets the curve \(\displaystyle y=\sqrt{3x}\) at one point is ...
5. A circle meets the line x + y = 3 at (2,1). It also meets the point (6,3). Its radius is ...

I have no idea how to do number 3, or even the meaning.

For number 4, I substituted x = 6 and y = 0 to the equation \(\displaystyle (x-a)^2+(y-b)^2=r^2\) and got \(\displaystyle 36-12a+a^2+b^2=36\).

For number 5, I substituted both coordinates to the circle equation and got 2a + b = 10.

Please help how to continue each number.

 

[MarkFL]

3.) I am thinking to let point $P$ be $(x,y)$. We are told then to let:

\(\displaystyle \frac{x-a}{y-b}\cdot\frac{x+a}{y+b}=k\)

Multiply through by $(y+b)(y-b)$ to get:

\(\displaystyle x^2-a^2=k(y^2-b^2)\)

Rearrange as:

\(\displaystyle x^2-ky^2=a^2-kb^2\)

So, what must $k$ be in order for this to be a circle?

 

[Monoxdifly]

Rearrange as:

\(\displaystyle x^2-ky^2=a^2-kb^2\)

So, what must $k$ be in order for this to be a circle?

-1?
That's the only way I can think to get rid of that -k from the coefficient of y.

 

[MarkFL]

-1?
That's the only way I can think to get rid of that -k from the coefficient of y.

Yes, the general form for a circle is:

\(\displaystyle (x-h)^2+(y-k)^2=r^2\)

This is a circle centered at $(h,k)$ with radius $r$. The $k$ here has no relation to the $k$ in the problem. So, we need the coefficient of $x^2$ and $y^2$ to be the same, which means $k$ must be $-1$, which also means the segments $\overline{PA}$ and $\overline{PB}$ will always be perpendicular, related to the fact that in a semicircle, any triangle whose base is on the diameter of the semicircle and its other vertex is on the circular portion of the semicircle, will be a right triangle.

 

[Monoxdifly]

Thank you. :)

What about number 4 and 5?

 

[MarkFL]

For problem 4, if the radius of the circle is $r$, then where must its center be? What relation must there be between the radius touching the parabolic curve and the line tangent to the parabola at that point?

 

[MarkFL]

If the circle touches the $x$-axis only at $(6,0)$, then the circle must be centered at $(6,r)$.

Now, let's let the circle touch the curve $y=\sqrt{3x}$ at $(a,\sqrt{3a})$. The slope of that curve at the tangent point is:

\(\displaystyle y'(a)=\frac{3}{2\sqrt{3a}}=\frac{1}{2}\sqrt{\frac{3}{a}}\)

And so the radius of the circle to the tangent point must have a slope perpendicular to the tangent line, so we may state:

\(\displaystyle \frac{r-\sqrt{3a}}{6-a}=-2\sqrt{\frac{a}{3}}\)

We also know the distance from the center of the circle to the tangent point is $r$, hence:

\(\displaystyle r^2=(6-a)^2+(r-\sqrt{3a})^2\)

If we expand this second equation, we obtain:

\(\displaystyle r^2=36-12a+a^2+r^2-2r\sqrt{3a}+3a\)

\(\displaystyle 0=36-9a+a^2-2r\sqrt{3a}\)

\(\displaystyle r=\frac{a^2-9a+36}{2\sqrt{3a}}\)

Substituting into the first equation, we find:

\(\displaystyle \frac{\dfrac{a^2-9a+36}{2\sqrt{3a}}-\sqrt{3a}}{6-a}=-2\sqrt{\frac{a}{3}}\)

Multiplying through by $2\sqrt{3a}$ we get:

\(\displaystyle \frac{a^2-15a+36}{6-a}=-4a\)

\(\displaystyle a^2-15a+36=4a^2-24a\)

\(\displaystyle 3a^a-9a-36=0\)

\(\displaystyle a^2-3a-12=0\)

Applying the quadratic formula, and discarding the negative root, we obtain:

\(\displaystyle a=\frac{3+\sqrt{57}}{2}\)

Hence:

\(\displaystyle r=(13-\sqrt{57})\sqrt{\frac{3}{2(13+\sqrt{57})}}\approx2.0551\)

 

[MarkFL]

For problem 5), let's let the equation of the circle be:

\(\displaystyle (x-h)^2+(y-k)^2=r^2\)

Now, we know the slope of the radius to the given line is $1$, thus:

\(\displaystyle k=h-1\)

We also know from the 2 given points on the circle that:

\(\displaystyle r^2=(2-h)^2+(1-k)^2=(6-h)^2+(3-k)^2\)

Expand:

\(\displaystyle 4-4h+h^2+1-2k+k^2=36-12h+h^2+9-6k+k^2\)

Simplify and solve for $k$:

\(\displaystyle k=10-2h\)

Thus:

\(\displaystyle h-1=10-2h\implies h=\frac{11}{3}\implies k=\frac{8}{3}\)

Hence:

\(\displaystyle r=\sqrt{\left(2-\frac{11}{3}\right)^2+\left(1-\frac{8}{3}\right)^2}=\frac{5}{3}\sqrt{2}\)