What is the radius of the circle through $(-1,1)$ and touching the lines $x\pm y=2?$

The lines $x+y=2$ and $x-y=2$ are perpendicular to each other and the circle is touching both the lines,these lines are tangents to the circle.Let points of tangency be $P$ and $Q$,let the center of the circle be $O$ and let the point where the lines $x\pm y=2$ meet be $R$.$OP$ is perpendicular to $PR$ and $OQ$ is perpendicular to $QR$,therefore $OPRQ$ is a square and the point $(-1,1)$ does not lie on any of the lines $x\pm y=2$.

But now i am stuck,how to solve further.Please help me.

  • $\begingroup$ Do you want a geometric solution or an analytic-geometry solution? $\endgroup$ – H. R. Oct 5 '15 at 14:41
  • 1
    $\begingroup$ Your beginning doesn't help you one bit. I would suggest you first draw the problem so you get what the problem is, because the picture in your head seems plain wrong :o $\endgroup$ – Inuyaki Oct 5 '15 at 14:45
  • $\begingroup$ I want analytic-geometry solution,if possible. $\endgroup$ – diya Oct 5 '15 at 14:46
  • $\begingroup$ The equation of a circle passing through $(x_1, y_1)$ and touching the line L is given by $(x-x_1)^2 + (y-y_1)^2 +\lambda L = 0$. Are you allowed to use this ? $\endgroup$ – Shailesh Oct 5 '15 at 14:50
  • 2
    $\begingroup$ Forget for the moment about the point (-1, 1) and try to understand where centers of all circles, touching both lines, must be $\endgroup$ – HEKTO Oct 5 '15 at 14:58

Let $(h,k)$ be the center of the required circle.

The perpendicular distance from $(h,k)$ to $x+y-2=0$

=The perpendicular distance from $(h,k)$ to $x-y-2=0$

=The distance between $(h,k)$ to $(-1,1)$.

The first two equality gives $|\frac{h+k-2}{\sqrt 2}|=|\frac{h-k-2}{\sqrt 2}|$, which will imply either $k=0$ or $h=2$

So, you have two cases,

Case,I $k=0$

From the seond and third equality gives,

$\frac{|h-2|}{\sqrt 2}=\sqrt{(h+1)^2+(1)^2}$ squaring on both sides,

or $ \frac{(h-2)^2}{2}=(h+1)^2+1$

or $h^2-4h+4=2h^2+4h+2+2$

or $h^2+8h=0$

0r $h=0,-8$

Radius$=\sqrt{(-8+1)^2+(1)^2}=5\sqrt 2 or \sqrt{(0+1)^2+(1)^2}=\sqrt 2$

Case-II, $h=2$

$\frac{|2+k-2|}{\sqrt 2}=\sqrt{(2+1)^2+(k-1)^2}$

squaring on both sides,

or, $k^2= 18+2k^2-4k+2$

or, $k^2-4k+30$

This doesn't have real solutions. So case-II is not valid.

  • $\begingroup$ Your approach is not completely analytic as you are using some geometric facts. So this solution is not completely analytic. $\endgroup$ – H. R. Oct 5 '15 at 15:42
  • $\begingroup$ @H.R. Could you care to explain what is meant by "analytic'? $\endgroup$ – Babai Oct 5 '15 at 16:39
  • $\begingroup$ Sorry for that, I just misunderstood some parts! :) My bad. $\endgroup$ – H. R. Oct 5 '15 at 16:53

there are two circles that will fit your questions. the plane is broken into four quadrants by the lines $x \pm y = 2$ that intersect at $B = (2, 0).$ the two circles are in the same quadrant as the point $A = (-1,1).$ since the center is on the bisector of the two tangents, the center has coordinate $O=(a, 0).$

the radius of the circle can be found in two ways:

(a) the center is a distance $\frac{|OB|}{\sqrt 2} = \frac{|a-2|}{\sqrt 2}$ from the tangents,

(b) is also $|OA| = \sqrt{(a+1)^2 + 1}$

equating the two you find that $a = 0, a= -8.$

  • $\begingroup$ What is meant by bisector of the tangents? The tangent is an infinite line. $\endgroup$ – Babai Oct 5 '15 at 16:45
  • $\begingroup$ @Babai, i meant one of the angle bisectors of the two lines in your question. $\endgroup$ – abel Oct 5 '15 at 16:59
  • $\begingroup$ okay, then it makes sense. $\endgroup$ – Babai Oct 5 '15 at 17:02

Here is another approach. The general equation for a circle is

$${\left( {x - a} \right)^2} + {\left( {y - b} \right)^2} = {R^2}$$

with the center at $(a,b)$ and the radius $R$. As the line $x+y=2$ is tangent to your circle, hence the line and the circle just have one intersection that you called point $P$, so the coordinates of $P$ is the solution to the nonlinear algebraic system

$$\left\{ \matrix{ {\left( {{x_P} - a} \right)^2} + {\left( {{y_P} - b} \right)^2} = {R^2} \hfill \cr {x_p} + {y_p} = 2 \hfill \cr} \right.$$

by assumption this system has exactly one solution. Considering ${y_p} = 2 - {x_p}$ and putting it into the first equation you may obtain a quadratic equation in terms of ${x_p}$. In order to force this quadratic equation have just one real root, it's discriminant must be zero. Carrying out the computations gives

$$\left\{ \matrix{ {\Delta _P} = 8{R^2} - 4{\left( {a + b - 2} \right)^2} = 0 \hfill \cr {x_p} = {a \over 2} - {b \over 2} + 1 \hfill \cr {y_p} = - {a \over 2} + {b \over 2} + 1 \hfill \cr} \right.$$

where the ${\Delta _P}$ was the discriminant of the aforementioned quadratic equation. You can repeat the same process for the point $Q$, the only intersection of the circle and the line $x-y=2$, which results in

$$\left\{ \matrix{ {\Delta _Q} = 8{R^2} - 4{\left( {a - b - 2} \right)^2} = 0 \hfill \cr {x_Q} = {a \over 2} + {b \over 2} + 1 \hfill \cr {y_Q} = {a \over 2} + {b \over 2} - 1 \hfill \cr} \right.$$

Now using the ${\Delta _P}$ and ${\Delta _Q}$ equations, you obtain

$$\left\{ \matrix{ \left| {a - b - 2} \right| = \left| {a + b - 2} \right| \hfill \cr R = {1 \over {\sqrt 2 }}\left| {a - b - 2} \right| = {1 \over {\sqrt 2 }}\left| {a + b - 2} \right| \hfill \cr} \right.$$

Then two cases are possible according to the first equation above.

Case 1. $b=0$
In this case the radius $R$ , the equation of circles, and coordinates of intersection points becomes

$$\left\{ \matrix{ R = {1 \over {\sqrt 2 }}\left| {a - 2} \right| \hfill \cr {\left( {x - a} \right)^2} + {y^2} = {1 \over 2}{\left( {a - 2} \right)^2} \hfill \cr \left\{ \matrix{ {x_p} = {a \over 2} + 1 \hfill \cr {y_p} = - {a \over 2} + 1 \hfill \cr} \right.\,\,\,,\left\{ \matrix{ {x_Q} = {a \over 2} + 1 \hfill \cr {y_Q} = {a \over 2} - 1 \hfill \cr} \right. \hfill \cr} \right.$$

Case2. $a=2$
In this case the radius $R$ , the equation of circles, and coordinates of intersection points becomes

$$\left\{ \matrix{ R = {1 \over {\sqrt 2 }}\left| b \right| \hfill \cr {\left( {x - 2} \right)^2} + {\left( {y - b} \right)^2} = {1 \over 2}{b^2} \hfill \cr \left\{ \matrix{ {x_p} = - {b \over 2} + 2 \hfill \cr {y_p} = {b \over 2} \hfill \cr} \right.\,\,\,,\left\{ \matrix{ {x_Q} = {b \over 2} + 2 \hfill \cr {y_Q} = {b \over 2} \hfill \cr} \right. \hfill \cr} \right.$$

So there are two family of circles, one having the center on the x-axis and one having the center on the line $x=2$.


The center of a circle tangent to both $x+y=2$ and $x-y=2$ lies on one of the angle bisectors: either the $x$-axis or the line $x=2$. If the circle passes through the point $(-1,1)$, the center must be on the $x$-axis where $x\lt2$.

enter image description here

If the center is at $(x,0)$, then the distance to either line $x\pm y=2$ is $$ \frac{\left|x\pm y-2\right|}{\sqrt2}=\frac{2-x}{\sqrt2} $$ so we need to solve the equation $$ \overbrace{(x+1)^2+1\vphantom{\frac{x^2}2}}^{\text{distance$^2$ from $(-1,1)$}}=\overbrace{\frac{(2-x)^2}2}^{\text{distance$^2$ from $x\pm y=2$}}\implies x=-8\text{ and }x=0 $$ Thus, there are two circles:

  1. Centered at $(0,0)$ with radius $\sqrt2$.

  2. Centered at $(-8,0)$ with radius $5\sqrt2$.

  • $\begingroup$ Would you please take a look at this question too. Thanks by the way. :) $\endgroup$ – H. R. Oct 5 '15 at 17:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.