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I have in fact solved this question. My process involved finding the distances of (h,k) ie. the center of the circle to the given lines and points, equating them to r (radius) and solving the three equations to obtain the three unknown quantities.

It had a lot of calculations. Can I use the fact that the given lines are perpendicular to each other to reduce my calculations? Is there any other, shorter method to solve solve?

My answer was $$(x-\frac 65)^2+(y-\frac{12}{5})^2=1$$ and $$(x-2)^2+(y-8)^2=25$$

Aditya
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4 Answers4

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Here's is another and easier method to the problem.

Let $X = 3x - 4y + 1$ and $Y = 4x + 3x - 7$, i.e transforming the coordinates such that lines become coordinate axis. (As the lines are perpendicular, they can be rotated to get coordinate axis.)

Notice that point $(0,0)$ in this XY system, corresponds to $(1,1)$ in xy system and the point $(2,3)$ in xy system corresponds to $(2,1)$ here.

Required circle is $(X-R)^2 + (Y - R)^2 = R^2$, where R is radius of circle and the circle passes through $(2,1)$. Solving will give you values of $R$ to be 1 and 5 and back substitution of $X$ and $Y$ will get you the desired circles.

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The observation that the two tangents are perpendicular to each other would reduce the problem considerably. Note that the center of the circle lies on the angle bisecting line between the two tangents. It passes the intersection of the two lines at $A(1,1)$ and is 45-degree steeper than that of the line $3x-4y-1=0$ whose slope is $\tan\theta=\frac34$. Therefore, the slope of the bisector line is

$$\tan(\theta + 45) = \frac{\tan\theta + 1}{1-\tan\theta} = \frac{\frac34+1}{1-\frac34} = 7$$

and, hence, its equation $y=7x-6$.

Let the center of the circle be $O(a,7a-6)$, its radius $r$, the given point $B(2,3)$ and the two tangential points $P$ and $Q$. Then, use the fact that $APOQ$ is a square with $AO$ as the diagonal, i.e. $AO = \sqrt2r=\sqrt2OB$, or

$$(a-1)^2+(7a-7)^2 = 2(a-2)^2+2(7a-9)^2$$

After similification,

$$5a^2-16a+12=0$$

Solve to obtain the center of the circle $(2,8)$ and $(\frac65,\frac{12}5)$.

Quanto
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Hint: there are two circles, one is $5$ times bigger than the other:

enter image description here

The idea is to start with any two convenient points, say, $B,D:\ |AB|=|BD|$, make a square $ABCD$ and a circle $\Gamma$ centered at $C$ with the radius $|CB|$. The centers of all the other circles, tangent to the given lines lie on the line $AC$. Then find the points, say, $K,L$ of intersection of $\Gamma$ with the line $AP$ and use $\frac{|AP|}{|AK|}$ and $\frac{|AP|}{|AL|}$ as the scaling factors to find the centers and radii of the two sought circles.

g.kov
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  • @amd: The hint is in the picture, using squares, the line $AP$ and similarity (scaling). – g.kov Feb 05 '20 at 20:23
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I would approach the problem in general as follows.

We are considering only the case in which the lines are incident. If they are parallel that's an easier case.

Let's call $P$ the point that the circle shall pass through.

a) Determine the crossing point $H$ of the two lines;

b) Take the vectors parallel to the lines and normalize them to obtain ${\bf t}_1, {\bf t}_2$.

c) Determine in which of the four angles between the lines the point $P$ is, by solving $a{\bf t}_1+b {\bf t}_2 = \vec{HP}$ wrt $a,b$:
if $a$ turns out to be negative redefine ${\bf t}_1 = -{\bf t}_1$ and same for $b$:
so $P$ lies in the angle within the positive direction of both the vectors.

d) Take the vector defining the angle bisector as ${\bf t}_1+ {\bf t}_2$ and normalize it to get $\bf t$.

e) Define a point on the bisector as $\vec{OH}+ s \, \bf t$;
its distance from the lines will be $s | \vec{HP} \times {\bf t} |$;
its distance from $P$ is $| \vec{OH}+ s \, {\bf t} -\vec{OP} |$.

f) Equate the two distances, determining two solutions for $s$, check that both are non-negative;
compute accordingly the centers and the radii of the two circles.

G Cab
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