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The eqn of circle which touches both the axes and straight line 4x+3y=6 in the first quadrant and lies below it is.

The eqn of circle which touches both the axes and straight line 4x+3y=6 in the first quadrant and lies below it is.

Grade:11

1 Answers

Rajat
213 Points
2 years ago
The given line cuts the co-ordinate Axes at (3/2, 0) and (0, 2). This implies the circle lies in the first quadrant. The circle is equidistant from both the Axes thus its equation is of the form: (x-a)2+(y-a)2=a2, where the distance of the centre of the circle from the axes and the given line ie. 4x+3y=6,  and the radius of the cirle is 'a' units. Coordinate of centre (a,a) and it is at a distance of a unit from given line. 
So, (4a+3a-6)/√(42+32) = a
   or, (7a-6)/5=a
   or, 7a-6=5a
   or, 2a=6
    So, a=3
 Therefore equation of the required circle is (x-3)2+(y-3)2=32

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