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Consider equation $x^2 + y^2 = h$ that touches the line $y=3x+2$ at some point $P$. Find the value of $h$

I know that $x^2 + y^2 = h$ is a circle with radius $\sqrt{h}$.

Also, since $y = 3x + 2 $ is a tangent, we know that the slope of the radius perpendicular to the tangent is $M_{OP}= -\frac{1}{3}$. I'm not sure how I can determine the value of $h$ though?

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10 Answers 10

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You're almost there! We want to find the equation of line $OP$. We know its slope, and (since it passes through the origin) we know its $y$-intercept, so its equation is: $$ y = \tfrac{-1}{3}x $$ We can now find the intersection point of the radius and tangent by solving the system of equations. Equating, we obtain: \begin{align*} \tfrac{-1}{3}x = y = 3x + 2 &\implies -x = 9x + 6 \\ &\implies -10x = 6 \\ &\implies x = \tfrac{-3}{5} \\ &\implies y = \tfrac{-1}{3} \cdot \tfrac{-3}{5} = \tfrac{1}{5} \end{align*} Thus, we conclude that: $$ h = x^2 + y^2 = (\tfrac{-3}{5})^2 + (\tfrac{1}{5})^2 = \tfrac{9 + 1}{25} = \boxed{\tfrac{2}{5}} $$

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So we have equation $x^2 + y^2 = r^2$, who’s geometrical representation would be a circle with the radius $r$. Now we also have an equation of a line represented by: $y = mx + c$ that touches the circle at point $P$. To find this point we will use the method of solving simultaneous equation where one is a quadratic and the other is linear so we will take the value of y and replace it in the first equation as follows:

$$x^2 + (3x + 2)^2 = h$$ $$x^2 + 9x^2 + 2*3x*2 + 4 = h$$ $$10x^2 + 12x + 4 = h$$ $$10x^2 + 12x + 4 - h = 0$$

so now we will have something like:

$$x^2 + (mx + c)^2 = r^2$$ $$x^2 + m^2x^2 + 2mcx + c^2 = r^2$$ $$(1 + m^2)x^2 + 2mcx + c^2 - r^2 = 0$$

since this equation is a quadratic equation in x, something like $(ax^2 + bx + c = 0)$ where we have:

$a = 10$; $b = 12$; $c = 4 - h$;

using the values in discriminant $(b^2 - 4ac)$:

$$b^2 - 4ac = (2mc)^2 - 4(1 + m^2)(c^2 - r^2)$$ $$= 4m^2c^2 - 4(c^2 + m^2c^2 - r^2 - m^2r^2)$$ $$ = 4m^2c^2 - 4c^2 - 4m^2c^2 + 4r^2 + 4m^2r^2$$ $$ = -4c^2 + 4r^2 + 4m^2r^2$$

if the discriminant is zero, our equation will have equal roots and the line intersects the circle in one single point which is our case also. So because this line is a tangent to the circle we know:

$$-4c^2 + 4r^2 + 4m^2r^2 = 0$$ $$4r^2 + 4m^2r^2 = 4c^2$$ $$c^2 = r^2 + m^2r^2$$

this will be the condition for a line to be tangent to the circle, so replacing with our values we will have the tangent:

$$(4 - h)^2 = h + 3^2h$$ $$h^2 - 8h + 16 = h + 9h$$ $$h^2 - 18h = -16$$ $$h(h - 18) = -16$$

since: $$(h-9)^2 = (h-9)(h-9)$$ $$h^2 - 18h + 81$$ $$h(h -18) + 81$$ we can write:

$$(h-9)^2 - 65 = 0$$ so:

$$h = 9 \pm \sqrt65$$

I hope this helps

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Hint:
What you are actually doing is trying to find the distance from $(0,0)$ to the line. Did this rephrasing of the problem help you?

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The condition requires that the equation $x^2+(3x+2)^2 - h = 10x^2+12x+4-h=0$ has a double root. Thus $\triangle' = 0\implies 6^2-10(4-h) = 0$. Can you find $h$ from this linear equation?

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Hint:

Take the system: $$ \begin{cases} x^2+y^2=h\\ y=3x+2 \end{cases} $$

whose solutions are the coordinates of the common points of the circle and the line. If the discriminant of this system is $=0$ it has a double solution, so the line is tangent to the circle.

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Hint

$h$ is fixed. Let $A=\{(x,y)\mid x^2+y^2=h\}$ and $B=\{(x,y)\mid y=3x+2\}$. $$(x,y)\in A\cap B\implies x^{2}+(3x+2)^2=h\implies ...\implies x=...$$

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The center of your circle is $(0\mid 0)\quad$.
The distance from the center to your line is $\frac2{\sqrt{10}}\quad$, which is equal to $\sqrt{h}\quad$.

$\sqrt{h}=\frac2{\sqrt{10}}\quad$, so $h=\frac4{10}=\frac25\quad$.

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Touching means $x^2 + y^2 = h$ and $y = 3x + 2 $ intersect along two coincident points, with two roots equal.

So eliminate $y$ between them and set the discriminant of quadratic in $x$ to zero, you get $h = \frac25$.

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There are couple of approaches already mentioned in other answers, but I will try to summarize them and add (probably unwarranted) generality. Also, I will describe approach using calculus.

If you are not interested in this generalities, please skip to the last paragraph.


We are given a line $y = ax + b$ and a circle centered $(p,q)$ given with $(x-p)^2+(y-q)^2 = h$ and we are looking for $h$ such that given line is tangent to the circle.

1. Calculus

We have function $y(x)$ for which relation $(x-p)^2+(y-q)^2 = h$ holds. If we differentiate the relation with respect to $x$ we get $$2(x-p)+2(y-q)y'=0\implies y'=-\frac{x-p}{y-q}$$ Now, since we are looking for a tangent given with $y = ax +b$, we are looking for a point $x$ such that $y'(x) = a$ and $y(x) = ax +b$. Thus we get $$a = -\frac{x-p}{ax + b - q}\implies x =\frac{p-ab+aq}{1+a^2}$$ and also $$y = ax + b = \frac{b + ap + a^2 q}{1+a^2}$$ Now, we get $$\boxed{h = (x-p)^2 + (y-q)^2 = \frac{(b+ap-q)^2}{1+a^2}}$$

2. Tangent of a circle is perpendicular to its radius

This is your original approach and is completely covered in the answer by Adriano.

Now, the line passing through center $(p,q)$ and is perpendicular to $y = ax + b$ is given by formula $$y - q = -\frac 1 a(x-p)$$ and we can find the intersection with the original line by solving linear system

\begin{align} y &= ax + b \\ y - q &= -\frac 1 a(x-p)\end{align}

to get $x = \frac{p - ab + aq}{a^2 + 1},\ y = \frac{b + ap + a^2 q}{a^2 + 1}$ and if we plug this into $(x-p)^2+(y-q)^2 = h$ we get $$\boxed{h = \frac{(b+ap-q)^2}{1+a^2}}$$

3. $h$ is the square of distance of the center to the line

We find this approach in answers by Senex Ægypti Parvi and pajonk.

We start with line $y = ax + b$ and point $(p,q)$. Square of (Euclidean) distance of any point on the line $y = ax + b$ to the point $(p,q)$ is given by

\begin{align} (x-p)^2 + (y-q)^2 &= (x-p)^2 +(ax+b-q)^2 \\ &= (1 + a^2) x^2 + 2( a b - p - a q) x+ (b^2 + p^2 - 2 b q + q^2)\end{align} and $h$ is given by minimal distance.

Now, remember that any quadratic $\alpha x^2 + \beta x + \gamma$ can be written in the form $$\alpha x^2 + \beta x + \gamma = \alpha(x + \frac\beta{ 2 \alpha})^2 + \gamma - \frac{\beta^2} {4 \alpha}\geq \gamma - \frac{\beta^2}{4\alpha}$$ and thus the minimum of quadratic polynomial ($\alpha > 0$) is given with $\gamma - \frac{\beta^2}{4\alpha}$ and if we set $\alpha = 1+ a^2,\ \beta = 2( a b - p - a q),\ \gamma = b^2 + p^2 - 2 b q + q^2$ we get $$\boxed{h = \gamma - \frac{\beta^2}{4\alpha} = \frac{(b+ap-q)^2}{1+a^2}}$$

4. Tangent is a line that intersects circle at exactly one point

I personally like this approach the best because it beautifully intertwines geometry and algebra. This is suggested in answers by cosmin, Emilio Novati and DeepSea.

We want to find intersections of the line $y = ax + b$ and circle $(x-p)^2+(y-q)^2 = h$ so we solve the appropriate system

\begin{align} (x-p)^2+(y-q)^2 &= h\\ y &= ax + b \end{align}

and by substituting $y = ax + b$ in the first equation we get $$(x-p)^2+(ax + b - q)^2 - h = 0$$ and analogously to the previous case we get quadratic equation $$ (1 + a^2) x^2 + 2( a b - p - a q) x+ (b^2 + p^2 - 2 b q + q^2 - h) = 0 $$

It is well known that there are 3 general cases for the solutions of quadratic equation (with real coefficients) depending on the discriminant $D$:

\begin{array}{ c | c | c } \textbf{discriminant} & \textbf{algebraic interpretation} & \textbf{geometric interpretation}\\ \hline D>0 & \text{there are 2 real solution} & \text{line intersects the circle at 2 points} \\ D=0 & \text{there is a double real solution} & \text{line intersects the circle at 1 point, i.e. it is a tangent} \\ D<0 & \text{there are 2 complex solutions} & \text{line does not intersect the circle } \end{array}

We are, thus, interested in case 2), i.e. we want the discriminant of our quadratic equation to be $0$, i.e.:

$$(2 a b - 2 p - 2 a q)^2 = 4 (1 + a^2) (-h + b^2 + p^2 - 2 b q + q^2) \iff \boxed{h = \frac{(b+ap-q)^2}{1+a^2}}$$


TL;DR

What we actually calculated is a well known formula from high school:

For a line $Ax + By + C = 0$, distance of point $(p,q)$ from the line is given by formula $$d = \frac{| Ap + Bq + C|}{\sqrt{A^2 + B^2}}$$

How to get the formula from our calculation of $h$? Well:

$$Ax + By + C = 0\implies y = -\frac AB x -\frac CB\implies h = \frac{(-C-Ap-Bq)^2}{B^2(1+\frac{A^2}{B^2})} = \frac{(Ap+Bq+C)^2}{A^2 + B^2}$$ Thus, probably the quickest way to solve your problem is

5. High school mathematics

Transform $y = 3x + 2$ into $3x - y + 2 = 0$, and put it in the above formula with $p = q = 0$ to get $$h = \frac{(3\cdot 0 - 1\cdot 0 + 2)^2}{3^2 + (-1)^2} = \frac 2 5$$

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The slope of the perpendicular radius is $-\dfrac 13$ and it passes through the center of the circle, $(0,0)$. So its equation is $y-0 = -\dfrac 13(x - 0)$, which simplifies to $y = -\dfrac 13x$. Avoiding fractions, we write this as $x = -3y$.

We compute the intersection

\begin{align} x &= -3y \\ y &= 3x + 2 \\ \hline y &= -9y + 2\\ y &= \dfrac 15 \\ x &= -\dfrac 35 \end{align}

So $h = x^2 + y^2 = \dfrac{10}{25} = \dfrac 25$

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