Question:
The equation of the circle which touches X-axis at (3, 0) and passes through (1,4) is given by
The equation of the circle which touches X-axis at (3, 0) and passes through (1,4) is given by
Options:
- x2 + y2 + 6x – 5y + 9 = 0
- x2 + y2 – 6x + 5y – 9 = 0
- x2 + y2 + 6x + 5y – 9 = 0
- x2 + y2 – 6x – 5y + 9 = 0
- x2 + y2 + 6x – 5y + 9 = 0
- x2 + y2 – 6x + 5y – 9 = 0
- x2 + y2 + 6x + 5y – 9 = 0
- x2 + y2 – 6x – 5y + 9 = 0
Correct Answer: a)
Explanation:
‘
‘ Let the equation of circle ha√ing centre (h, k) and radius a is (x-h)2 +(y- K)2 = a2
1 Here, centre (h, k) =(3,k) anda=k
‘ Now, by distance formula,
hon a
Y!
(1./ 1
D
1 i 60 X
y
‘ Hence, the required equation of circle is (z — 3)? + (y — 3)? — (3)?
‘ =x2+y?—62—5y+9=0 ‘
‘
‘ Let the equation of circle ha√ing centre (h, k) and radius a is (x-h)2 +(y- K)2 = a2
1 Here, centre (h, k) =(3,k) anda=k
‘ Now, by distance formula,
hon a
Y!
(1./ 1
D
1 i 60 X
y
‘ Hence, the required equation of circle is (z — 3)? + (y — 3)? — (3)?
‘ =x2+y?—62—5y+9=0 ‘