Find equation of circle(s) tangent to $2x − y − 5 = 0$ and $2x + y − 7 = 0$ passing through $(10, −3)$.

Question

Find equation of circle(s) tangent to $2x − y − 5 = 0$ and $2x + y − 7 = 0$ passing through $(10, −3)$.

I tried using Angle bisector but couldn't get the solution

Answer 1

Note that bisectors will be parallel to axis $y$ and axis $x$ as slopes of tangent lines are opposite. The tangents intersect at $2x-5=-2x+7 \implies x=3, y=1$. Looking at the point $(10,-3)$ we can deduct that the center of the circle is on the line $y=1$ (this bisector goes thru the East quadrant where the point is located).
The distance from the center $(x_c,1)$ to point $(10,-3)$ is $r=\sqrt{(x_c-10)^2+(1+3)^2}=\sqrt{(x_c-10)^2+16}$.

The distance from the center to one of the lines: $r=\frac{|2x_c-1-5|}{\sqrt{4+1}}$.

$$\frac{|2x_c-6|}{\sqrt{5}}=\sqrt{(x_c-10)^2+16}$$

Squaring both sides: $(2x_c-6)^2=5(x_c-10)^2+80$ which will produce $x_c=8$ or $x_c=68$. Which leads to two equations: $$(x-8)^2+(y-1)^2=(8-10)^2+16=20$$ and $$(x-68)^2+(y-1)^2=(68-10)^2+16=3380$$

Answer 2

This is more geometric, by finding the locus of the center of the desired circle:

Let $C(O)$ be the desired circle, and $\ell_1: 2x - y - 5 = 0$, $\ell_2: 2x + y - 7 = 0$, and $A = (10, -3)$. Then:

• $C$ is tangent to $\ell_1$ and $\ell_2$, so $O$ lies on the bisectors of the angles between $\ell_1$ and $\ell_2$. (The equations of these two bisectors can be achieved from $\color{purple}{\dfrac{|2x - y - 5|}{\sqrt{5}} = \dfrac{|2x + y - 7|}{\sqrt{5}} \implies x = 3\ \text{or}\ y = 1}$.)
• The distance between $O$ and $\ell_1$ equals the distance between $O$ and $A$. So $O$ lies on a parabola with focus $A$ and directrix $\ell_1$. (Its equation is $\color{blue}{\sqrt{(x - 10)^2 + (y + 3)^2} = \dfrac{|2x - y - 5|}{\sqrt{5}} \implies x^2 + 4xy + 4y^2 - 80x + 20y + 520 = 0}$.)
• The distance between $O$ and $\ell_2$ equals the distance between $O$ and $A$. So $O$ lies on a parabola with focus $A$ and directrix $\ell_2$. (Its equation is $\color{red}{\sqrt{(x - 10)^2 + (y + 3)^2} = \dfrac{|2x + y - 7|}{\sqrt{5}} \implies x^2 - 4xy + 4y^2 - 72x + 44y + 496 = 0}$.)

The parabolas don't intersect the line $x = 3$. So by putting $y = 1$ in the equation of one of the parabolas, we get $x^2 - 76x + 544 = (x - 8)(x - 68) = 0$. So $O = (8, 1)$ or $O = (68, 1)$.

Answer 3

The centers of the circles are equidistant from both lines, so if $(C_1, C_2)$ is a center then

$\dfrac{ |2 C_1 - C_2 - 5 |}{\sqrt{2^2 + 1^2}} = \dfrac{ | 2C_1 + C_2 - 7 |}{\sqrt{2^2 + 1^2}}$

which simplifies to

$ |2 C_1 - C_2 - 5 |= | 2C_1 + C_2 - 7 | $

Squaring both sides

$ 4C_1^2 + C_2^2 + 25 - 4 C_1 C_2 - 20 C_1 + 10 C_2 = 4 C_1 + C_2^2 + 49 + 4 C_1 C_2 - 28 C_1 - 14 C_2 $

And this simplifies to

$ 8 C_1 C_2 - 8 C_1 - 24 C_2 + 24 = 0 $

And further to

$ C_1 C_2 - C_1 - 3 C_2 + 3 = 0 \hspace{30pt} (1)$

The second condition we have comes from the equation of the circle

$ (x - C_1)^2 + (y - C_2)^2 = r^2 = \left( \dfrac{ |2 C_1 - C_2 - 5 |}{\sqrt{2^2 + 1^2}} \right)^2 = \dfrac{1}{5} \left( 4C_1^2 + C_2^2 + 25 - 4 C_1 C_2 - 20 C_1 + 10 C_2 \right) $

The point $(10,-3)$ is on the circle, so

$ 5 \left( (10 - C_1)^2 + (-3 - C_2)^2 \right) = 4C_1^2 + C_2^2 + 25 - 4 C_1 C_2 - 20 C_1 + 10 C_2 $

Simplifying the left hand side, we end up with

$ C_1^2 + 4 C_1 C_2 + 4 C_2^2 -80 C1 + 20 C_2 + 520 = 0 \hspace{30pt} (2) $

Solving $(1) \ \& \ (2)$ we obtain only TWO solutions

• Circle $1$: Center $ (8, 1)$ , Radius $= \sqrt{20} $
• Circle $2$: Center $(68, 1)$ , Radius $ =\sqrt{3380} $

Answer 4

Let's assume a circle $(x-a)^2+(y-b)^2=r^2$ is tangent to $2x−y−5=0$, then by eliminating y we can get a quadratic equation about x:$$(x-a)^2+(2x-b-5)^2=r^2$$

x should have only one root so the discriminant should be zero. That is: $$-16a^2+16ab+80a-4b^2-40b+20r^2-100=0\quad\text{(1)}$$

Analogously, when the circle is tangent to $2x+y−7=0$, we get: $$16a^2-16ab+112a-4b^2+56b+20r^2-196=0\quad\text{(2)}$$

When the circle passes through $(10,-3)$, we get: $$(10-a)^2+(-3-b)^2=r^2\quad\text{(3)}$$

The real solutions of a, b and r for Eq (1), (2) and (3) are: $$\begin{cases}a=8\\b=1\\r=\pm2\sqrt{5}\end{cases}$$ and $$\begin{cases}a=68\\b=1\\r=\pm26\sqrt{5}\end{cases}$$

So their are two circles fit the condition: $$(x-8)^2+(y-1)^2=20$$ and $$(x-68)^2+(y-1)^2=3380$$