Q80 / 180 · COMEDK 2021 · Mathematics
What will be the equation of the circle whose centre is $(1,2)$ and which passes through the point $(4,6)$ ?
- Correct answerA$x^{2}+y^{2}-2 x-4 y-20=0$
- B$x^{2}+y^{2}+2 x+4 y-20=0$
- C$x^{2}-y^{2}-2 x-4 y+20=0$
- D$x^{2}-y^{2}+2 x-4 y-20=0$
Solution
The radius $r$ of the circle is the distance between the centre $(1, 2)$ and the point $(4, 6)$ on the circle.
$r = \sqrt{(4 - 1)^{2} + (6 - 2)^{2}}$
$r = \sqrt{3^{2} + 4^{2}} = \sqrt{9 + 16} = \sqrt{25} = 5$
The equation of a circle with centre $(h, k)$ and radius $r$ is given by $(x - h)^{2} + (y - k)^{2} = r^{2}$.
Substituting $(h, k) = (1, 2)$ and $r = 5$:
$(x - 1)^{2} + (y - 2)^{2} = 5^{2}$
$x^{2} - 2x + 1 + y^{2} - 4y + 4 = 25$
$x^{2} + y^{2} - 2x - 4y + 5 - 25 = 0$
$x^{2} + y^{2} - 2x - 4y - 20 = 0$
Answer: $x^{2}+y^{2}-2 x-4 y-20=0$
$r = \sqrt{(4 - 1)^{2} + (6 - 2)^{2}}$
$r = \sqrt{3^{2} + 4^{2}} = \sqrt{9 + 16} = \sqrt{25} = 5$
The equation of a circle with centre $(h, k)$ and radius $r$ is given by $(x - h)^{2} + (y - k)^{2} = r^{2}$.
Substituting $(h, k) = (1, 2)$ and $r = 5$:
$(x - 1)^{2} + (y - 2)^{2} = 5^{2}$
$x^{2} - 2x + 1 + y^{2} - 4y + 4 = 25$
$x^{2} + y^{2} - 2x - 4y + 5 - 25 = 0$
$x^{2} + y^{2} - 2x - 4y - 20 = 0$
Answer: $x^{2}+y^{2}-2 x-4 y-20=0$
Hence, the correct answer is option: 'A'.