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Question

Question: The equation of the circle of radius 5 and touching the coordinate axes in third quadrant is....

The equation of the circle of radius 5 and touching the coordinate axes in third quadrant is.

A

(x5)2+(y+5)2=25( x - 5 ) ^ { 2 } + ( y + 5 ) ^ { 2 } = 25

B

(x+4)2+(y+4)2=25( x + 4 ) ^ { 2 } + ( y + 4 ) ^ { 2 } = 25

C

(x+6)2+(y+6)2=25( x + 6 ) ^ { 2 } + ( y + 6 ) ^ { 2 } = 25

D

(x+5)2+(y+5)2=25( x + 5 ) ^ { 2 } + ( y + 5 ) ^ { 2 } = 25

Answer

(x+5)2+(y+5)2=25( x + 5 ) ^ { 2 } + ( y + 5 ) ^ { 2 } = 25

Explanation

Solution

Since circle touches the co-ordinate axes in III quadrant.

\therefore Radius =h=k= - h = - k . Hence h=k=5h = k = - 5

\therefore Equation of circle is (x+5)2+(y+5)2=25( x + 5 ) ^ { 2 } + ( y + 5 ) ^ { 2 } = 25