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

Question

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

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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.

Answer 1

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.

Answer 2

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 $a = 0, a= -8.$

Answer 3

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 $x=2$.

Answer 4

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 $x\lt2$.

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 $5\sqrt2$.