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# Anyone help !!!

I found this problem was very difficult :(. Can anyone help me solve this? Here is the problem:

Given a parabola $$y = x^2 - 8x + 12$$. Find the ratio of two inscribed circle on the parabola, both bordered by the parabola and the $$x$$ axis. But one circle is over $$x$$ axis and the other one is under $$x$$ axis.

Note by Muh Amin
1 year, 3 months ago

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Both the circles touch X axis at (4,0) by symmetry.We can interpret this as follows -these two circles are the members of family of circles that pass through through the intersection of line y=0 and the circle with centre (4,0) and radius 0.We know that S+ku=0 represents family of circles passing through intersection of circle S=0 and line u=0 where k is arbitrary parameter.Now solve the equation of circle and parabola to obtain a quadratic equation in y.Since the parabola and circle touch each other set the discriminant of the quadratic as 0.This will give two values of k namely 3 and -5.The two values of k give us the two equations of the circles. · 1 year, 1 month ago

the length of the circumference O1 = 15.71(approximately)

the length of the circumference O2 = 9.43(approximately)

Co1/Co2 = 1.7

· 1 year, 2 months ago

Is there no other 'common' way to solve this problem? I mean without using software? I just want to know how to solve this problem. But firstly, thank you so much for your help. · 1 year, 2 months ago

I don't know English very well, but I'll try to help you :

First of all let's rewrite our function in this form (it will be the same result at the end since I am using Translation (geometry)):

$$f(x)=x^{2}-4$$

Then we must find the coordinates of the vertex of our new parabola and roots of $$x^{2}-4=0$$

Here it is :

1) coordinates of the vertex : $$x=0$$ and $$y=-4$$ ,so : $$(0;-4)$$

2) roots : $$x = \pm2$$ , also we have two points : $$(-2;0)$$ and $$(2;0)$$

So, now we should define some points (take a look at my picture) : $$A (0;-4)$$ , $$B (-2;0)$$ , $$C (2;0)$$ also let's suppose that $$O1$$ and $$O2$$ are the centres of our circles, where $$O1(red) (0;y)$$ and $$O2 (0;y1)$$ .

Let's take a look at triangle $$ABC$$ :

$$AB=AC = \sqrt { 20 }$$(by Pythagorean theorem) and $${ S }_{ \Delta ABC } = 8$$

Let's draw circumcircle $$O3$$(orange) of the triangle $$ABC$$ and find it's radius : $$R=\frac { AB*BC*CA }{ 4{ S }_{ ABC } }$$, where $$R= 2.5$$ .

So, the centre of $$O3$$ is obviously point $$(0;-1.5)$$

/**********************************************************

Let's suppose that point $$(0;-1.5)$$ is also the centre of circle $$O1$$. It means, that $$R(O1)=1.5$$.

To prove that point $$(0;-1.5)$$ is the centre of $$O1$$ we must prove that line segment $$LO1=1.5$$.

Straight line $$LO1$$ is a $$normal$$- line that is perpendicular to our parabola in point $$L$$.

So, here it's function : $$f(x)=-\frac { 1 }{ 2{ x }_{ 0 } } (x-{ x }_{ 0 })+({ { x }_{ 0 } }^{ 2 }-4)$$ ($${ x }_{ 0 }$$ is the touch abscissa(point $$L$$) of the parabola and circle $$O1$$)

Let's find $${ x }_{ 0 }$$. Under our assumption point $$O1$$ has the following coordinates $$(0;-1.5)$$. Now we must put them in our $$normal$$ function.

$$-\frac { 1 }{ 2{ x }_{ 0 } } (0-{ x }_{ 0 })+({ { x }_{ 0 } }^{ 2 }-4)=-1.5$$

Where $${ x }_{ 0 }=-1.414213562$$. Now we must put $${ x }_{ 0 }$$ in the parabola function.

$$y={(-1.414213562) }^{ 2 }-4$$ , $$y=-2$$. Ordinate of point $$L=-2$$. So, $$L=(-1.414213562;-2)$$

Now, we should check if $$|LO1|=|O1D|=1.5$$

$$|LO1|=\sqrt { { (0-(-1.414213562)) }^{ 2 }+{ (-1.5-(-2)) }^{ 2 } } = 1.5$$.

Our assumptions were confirmed =) . So point $$(0;-1.5)$$ is the centre of circle $$O1$$. And of course $$|O1D|={ R }_{ O1 }=1.5$$

/**********************************************************

Now, the next part of my horrible solution.

It's much easier to find the centre of $$O2$$.

Let's assume that the straight line $$KB$$ coincides with our parabola ($$y\in (0;6)$$) and tangent line to the parabola at point $$B(-2;0)$$

So, we can find $$\tan { (\angle KBD) }$$

Firstly, let's find the tangent line function : it's $$f(x)'=({ x }^{ 2 }-4)'=2x$$ then put abscissa of point $$B$$ in this function.

So, $$f'(-2)=-4=\tan { \angle } KBD \Rightarrow { \angle } KBD\approx { 104 }^{ \circ }$$

Since $$KB$$ and $$BD$$ are tangents to the circle $$O2$$ $$\Rightarrow$$ $$KB=BD$$. Also $$KO2=O2D$$ $$\Rightarrow$$ $$O2B$$-bisector and median $$\Rightarrow$$ $$O2BD\approx { 52 }^{ \circ }$$

$$\tan { \angle } O2BD\approx 1.28$$ $$\Rightarrow$$ $$1.28=\frac { O2D }{ DB } =\frac { O2D }{ 2 }$$ $$\Rightarrow$$ $$O2D=2.5$$

$$O2=(0;2.5)$$ $$\Rightarrow$$ $${ R }_{ O2 }=2.5$$

/**********************************************************

And of course the last step :

$$\frac { { R }_{ O1 } }{ { R }_{ O2 } } =0.6$$

That's it $$=)$$ · 1 year, 2 months ago

Thank you very much Fatum. My english is not very good also :) · 1 year, 2 months ago

But why exactly this problem, why did you want to solve it? · 1 year, 2 months ago

I just randomly solve problem from my book. And from those problems, I just can't solve this one · 1 year, 2 months ago

What book? · 1 year, 2 months ago