Q. 193.7( 3 Votes )

The equation of t A. 4

B. 2

C. 3

D. 6

Answer :

Given that the equation of the circle which touches the axes of coordinates and the line and whose centres lie in the first quadrant is x2 + y2 - 2cx - 2cy + c2 = 0. We need to find the value of c.



We know that for a circle x2 + y2 + 2ax + 2by + c = 0


Centre = (- a, - b)


Radius =


For x2 + y2 - 2cx - 2cy + c2 = 0


Centre(C) =


C = (c,c)


Radius(r) =



r = c


We have touching the circle.


We know that the perpendicular distance between the circle and centre is equal to the radius of the circle.


We know that the perpendicular distance between from the point (x1,y1) to the line ax + by + c = 0 is



.







c2 - 7c + 6 = 0


c2 - 6c - c + 6 = 0


c(c - 6) - 1(c - 6) = 0


(c - 1)(c - 6) = 0


c - 1 = 0 or c - 6 = 0


c = 1 or c = 6


The correct option is (d).

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