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I have already found out the separate eqns of the pair of straight lines they are $x+y-1=0$ and $x-y+1=0$ and also their point of intersection of these two i.e. $(0,1)$.

Martund
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SUBHAM GHOSH
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  • Is x2 supposed to be $x^2?$ They are not the same. If you have a pair of intersecting lines there are many circles that touch them both. Please check the question. – Ross Millikan Nov 26 '19 at 04:17
  • @RossMillikan, yes that is supposed to be $x^2$, which is clear when we multiply the equation of lines he obtained. – Martund Nov 26 '19 at 04:19
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    @Martund: I feel people should make an effort to get the question right. Beyond what you found, there are many circles tangent to both lines, not just one. – Ross Millikan Nov 26 '19 at 04:21
  • As RossMillikan noted, some data is missing. Two intersecting lines divide coordinate plane into four parts, each part contains infinitely many circles touching both lines. – Vasili Nov 26 '19 at 04:27
  • Compare with https://math.stackexchange.com/questions/1484566/the-equation-of-a-pair-of-tangents-to-a-circle-from-a-point – lab bhattacharjee Nov 26 '19 at 04:29
  • Insufficient condition if you want to find unique circle . Because there will be infinite circles possible touching these two lines.Center of all such circles will lie on angling bisectors of to lines I.e on lines y=1 and another infinite circles with center on x=0 . – mathsdiscussion.com Nov 26 '19 at 04:38

1 Answers1

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If you mean the equation of circle tangent to lines $x+y-1=0$ and $x-y+1=0$, note that the lines are symmetry about y axis , so it's center has a coordinate like $C(0, a); [1<a<∞] and [-∞<a<1]$ .C is the intersection of a line perpendicular to one of these lines and y axis. Now take an arbitrary point $A(x_A, y_A)$ on one of these lines ; find the equation of perpendicular and hence a. $R=AC$ is the radius of the circle and the equation of a circle tangent at line at point $A(x_A, y_A)$ is:

$x^2+(y-a)^2=R^2$

$R^2=(y_A-a)^2+ x^2_A$

That shows there can be many circle tangent to these line depending on values of a and R.

The two lines are also symmetry about line $y=1$.That is the center of circles can have coordinate like $O(x_O=b , 1)$. You can follow the same method to find center O for any arbitrary point like $A(x_A, y_A)$ on each line. The equation of circle will be:

$(x-b)^2+(y-1)^2=R^2$

Where R can be found as:

$R^2=AO^2= (x_A-b)^2+(y_A-1)^2$

And b can be found by solving a system of equation resulted from intersection of a perpendicular from A and line $y=1$.

sirous
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