If pair of tangents to a circle in the first quadrant is $6x^2-5xy+y^2=0$ and if one point of contact is $(1,2)$, find the radius.

Question

The tangents are $2x-y=0$ and $3x-y=0$. Let the radius be $r$ and centre be $(h,k)$

$$r=\frac{|3h-k|}{\sqrt {10}}$$ $$r=\frac{|2h-k|}{\sqrt 5}$$ $$(h-1)^2+(k-2)^2=r^2$$

I invested a considerable amount of effort in solving this equations, but to no result. My method was to square all terms to avoid the modulus, but that complicated things adding the $hk$ term. This also led me to believe that there must be a better way to solve this. Can I know how this problem must be approached?

The radius is the distance from $(1,2)$ to the appropriate angle bisector. — amd, Apr 14 '20 at 17:21
Of the two tangents, of course. The centers of all of the circles to which they are mutually tangent lie on the two bisectors. — amd, Apr 14 '20 at 17:26

Matteo · Accepted Answer · 2020-04-14T18:59:33.037

1

The perpendicular to $y=2x$ throught $(1,2)$ is: $$y=-\frac{1}{2}x+\frac{5}{2}$$ The line bisector of the two lines $y=2x$ and $y=3x$ is: $$y=(\sqrt2+1)x$$ The other bisector line is: $$y=(\sqrt2-1)x$$ but in this case the circunference wouldn't be tangent to either $y=2x$ and $y=3x$ lines.

Now, we have to inresect these two lines, or: $$(\sqrt2+1)x=-\frac{1}{2}x+\frac{3}{2} \leftrightarrow x=5(3-2\sqrt2) \land y=-5+5\sqrt2$$ From this we can compute: $$r=\sqrt{(14-10\sqrt2)^2+(-7+5\sqrt2)^2}=\sqrt{(5\sqrt{10}-7\sqrt5)^2}=5\sqrt{10}-7\sqrt5$$

Note that we pass from $(14-10\sqrt2)^2+(-7+5\sqrt2)^2$ to $(5\sqrt{10}-7\sqrt5)^2$, simply calculating the first sum and then solve for $a,b$ this system: $$\left\{\begin{matrix} a^2+b^2=495 \\ 2ab=-350\sqrt2 \end{matrix}\right.$$ And the solutions are: $$a=\pm5\sqrt{10} \land y=\mp7\sqrt{5}$$

edited Apr 14 '20 at 18:59

answered Apr 14 '20 at 17:33

Matteo

6,581

Actually, I got $y=5(\sqrt 2-1)$ – Aditya Apr 14 '20 at 17:49
@Aditya: also I. – Matteo Apr 14 '20 at 17:57
Okay, so solving with that, I get $r=5\sqrt {12-8\sqrt 2}$. But $r$ is of the form $\sqrt {\alpha}-\sqrt {\beta}$, where $\alpha-\beta=5$ – Aditya Apr 14 '20 at 18:00
@Aditya: here my solution. – Matteo Apr 14 '20 at 18:06
It might be good to mention why you chose the particular bisector of the two that you did. – amd Apr 14 '20 at 18:27
I didn’t understand the part where you factorized them. What are $a$ and $b$ supposed to represent – Aditya Apr 15 '20 at 06:34
@Aditya: We suppose $(14-10\sqrt2)^2+(-7+5\sqrt2)^2$ can be represented as: $$(a+b)^2$$so we have: $$\left{\begin{matrix} a^2+b^2=495 \ 2ab=-350\sqrt2 \end{matrix}\right.$$ – Matteo Apr 15 '20 at 06:54

score 0 · Answer 2 · answered Apr 14 '20 at 17:27

0

The radius is equal to the distance from $(1,2)$ to the appropriate angle bisector of the two tangent lines. These bisectors are $${2x-y\over\sqrt5}=\pm{3x-y\over\sqrt{10}}.$$ Choose the one for which the circle will be in the first quadrant.

answered Apr 14 '20 at 17:27

amd

53,693

Solving it, I got the radius as $r=\frac{1}{\sqrt {20+14\sqrt 2}}$, and upon simplification ended up being $\frac{\sqrt{10-7\sqrt 2}}{2}$, but the answer is in the form of $\sqrt {\alpha}+\sqrt {\beta}$ – Aditya Apr 14 '20 at 17:37
@Aditya If you want to reverse-engineer the given solution, you’ll have to try to guess how $\alpha$ and $\beta$ relate to the values that you have. Perhaps you’re not meant to factor the original equation and work directly with that somehow. I’m not even going to hazard a guess without knowing $\alpha$ and $\beta$, though. – amd Apr 14 '20 at 18:46
I don’t know what alpha and beta are, I just know their difference. That’s the answer – Aditya Apr 15 '20 at 06:29

If pair of tangents to a circle in the first quadrant is $6x^2-5xy+y^2=0$ and if one point of contact is $(1,2)$, find the radius.

2 Answers2