Q74 / 180 · COMEDK 2021 · Mathematics
What will be the equation of circle whose centre is $(1,2)$ and touches $X$ - axis?
- Correct answerA$x^{2}+y^{2}-2 x-4 y+1=0$
- B$x^{2}-y^{2}+2 x+4 y+1=0$
- C$x^{2}+y^{2}+2 x-4 y-1=0$
- D$x^{2}+y^{2}+2 x+4 y-1=0$
Solution
The centre of the circle is given as $(h, k) = (1, 2)$.
Since the circle touches the X-axis, the radius $r$ of the circle is equal to the absolute value of the y-coordinate of the centre. Thus, $r = |k| = |2| = 2$.
The standard equation of a circle with centre $(h, k)$ and radius $r$ is $(x - h)^{2} + (y - k)^{2} = r^{2}$.
Substituting the values $h = 1$, $k = 2$, and $r = 2$ into the equation:
$(x - 1)^{2} + (y - 2)^{2} = 2^{2}$
Expanding the terms:
$(x^{2} - 2x + 1) + (y^{2} - 4y + 4) = 4$
Simplifying the equation:
$x^{2} + y^{2} - 2x - 4y + 5 = 4$
$x^{2} + y^{2} - 2x - 4y + 1 = 0$
Answer: $x^{2}+y^{2}-2 x-4 y+1=0$
Since the circle touches the X-axis, the radius $r$ of the circle is equal to the absolute value of the y-coordinate of the centre. Thus, $r = |k| = |2| = 2$.
The standard equation of a circle with centre $(h, k)$ and radius $r$ is $(x - h)^{2} + (y - k)^{2} = r^{2}$.
Substituting the values $h = 1$, $k = 2$, and $r = 2$ into the equation:
$(x - 1)^{2} + (y - 2)^{2} = 2^{2}$
Expanding the terms:
$(x^{2} - 2x + 1) + (y^{2} - 4y + 4) = 4$
Simplifying the equation:
$x^{2} + y^{2} - 2x - 4y + 5 = 4$
$x^{2} + y^{2} - 2x - 4y + 1 = 0$
Answer: $x^{2}+y^{2}-2 x-4 y+1=0$
Hence, the correct answer is option: 'A'.