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Question

The equation of a circle with radius 5 and touching both the coordinate axes is
(a) x2 + y2 ± 10x ± 10y + 5 = 0
(b) x2 + y2 ± 10x ± 10y = 0
(c) x2 + y2 ± 10x ± 10y + 25 = 0
(d) x2 + y2 ± 10x ± 10y + 51 = 0

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Solution

(c) x2 + y2 ± 10x ± 10y + 25 = 0

Case I: If the circle lies in the first quadrant:

The equation of a circle that touches both the coordinate axes and has radius a is x2+y2-2ax-2ay+a2=0.

The given radius of the circle is 5 units, i.e. a=5.

Thus, the equation of the circle is x2+y2-10x-10y+25=0.

Case II: If the circle lies in the second quadrant:

The equation of a circle that touches both the coordinate axes and has radius a is x2+y2+2ax-2ay+a2=0.

The given radius of the circle is 5 units, i.e. a=5.

Thus, the equation of the circle is x2+y2+10x-10y+25=0.

Case III: If the circle lies in the third quadrant:

The equation of a circle that touches both the coordinate axes and has radius a is x2+y2+2ax+2ay+a2=0.

The given radius of the circle is 5 units, i.e. a=5.

Thus, the equation of the circle is x2+y2+10x+10y+25=0.

Case IV: If the circle lies in the fourth quadrant:

The equation of a circle that touches both the coordinate axes and has radius a is x2+y2-2ax+2ay+a2=0.

The given radius of the circle is 5 units, i.e. a=5.

Thus, the equation of the circle is x2+y2-10x+10y+25=0.


Hence, the required equation of the circle is x2 + y2 ± 10x ± 10y + 25 = 0.

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