Question:
What is the equation of the circle which touches both the axes in the first quadrant and the line y – 2 = 0?
What is the equation of the circle which touches both the axes in the first quadrant and the line y – 2 = 0?
Options:
- x2 + y2 – 2x – 2y – 1 = 0
- x2 + y2 + 2x + 2y + 1 = 0
- x2 + y2 – 4x – 4y + 4 = 0
- x2 + y2 – 2x – 2y + 1 = 0
- x2 + y2 – 2x – 2y – 1 = 0
- x2 + y2 + 2x + 2y + 1 = 0
- x2 + y2 – 4x – 4y + 4 = 0
- x2 + y2 – 2x – 2y + 1 = 0
Correct Answer: d)
Explanation:
‘ The circle which touches both the axes in the first quadrant and the line y — 2 = 0 is shown
1 in the figure.
Y,
=2
¥
7
10
The diameter of the circle is 2.
‘ .. Radius = 1 and Centre of circle is (1, 1).
‘ .-. Equation of circle is ‘
(x—12 + (y-12=1 H
‘ Sx2 4+y?—2x-2y+1=0
163.
1 da?+y?—6z—5y+9=0
‘
‘ The circle which touches both the axes in the first quadrant and the line y — 2 = 0 is shown
1 in the figure.
Y,
=2
¥
7
10
The diameter of the circle is 2.
‘ .. Radius = 1 and Centre of circle is (1, 1).
‘ .-. Equation of circle is ‘
(x—12 + (y-12=1 H
‘ Sx2 4+y?—2x-2y+1=0
163.
1 da?+y?—6z—5y+9=0
‘