if two circles each of radius 5 unit touch each other at (1,2) and the equation of their common tangent is 4x+3y=10,then the equation of the circle, a portion of which lies in all the quadrants is:
(a)x2+y2-10x-10y+25=0
(b)x2+y2+6x+2y-15=0
(c)x2+y2+2x+6y-15=0
(d)x2+y2+10x+10y+25=0
-
UP 0 DOWN 0 0 1
1 Answers
21Solution 1:
Since the required circle passes through all the quadrants the origin lies within the circle.
Let the circle be S= x2 + y2 + 2gx +2fy + c=0
hence we get c<0.
Now the center of the circles lie on the line perpendicular to the tangent.
Then using polar form of the equation of this line we know that the centre is 5 units away from(1,2).
Hence we find the center of the two circles.
now replacing the centers as (-g,-f) in S find the value of c.
the center for which c<0 gives the answer,
Solution 2:
For such questions in jee we can find the answer using options. since the circle lies in all quadrants c<0 in the equation of circle x2 + y2 + 2gx +2fy + c=0
so options (a) and (d) are eliminated.
now only option (b) satisfies the point (1,2).