
The area of the circle whose centre is at $(1,2)$ and which passes through the point $(4,6)$ is
1) $5 \pi$
2) $10 \pi$
3) $25 \pi$
4) None of these
Answer
163.8k+ views
Hint: We know that the area of the circle is the area included inside the circumference or perimeter of the circle.$A=\pi r^{2}$, where $r$ is the circle's radius, is the formula for calculating a circle's area. with the given equation we can find the area of the given circle.
Complete step by step Solution:
Let $(h, k)$ be the centre of a circle with radius $r$.
Thus, its equation will be $(x-h)^{2}+(y-k)^{2}=r^{2}$
The location of a circle in the Cartesian plane is represented by a circle equation. We can write the equation for a circle if we know the location of the circle's centre and how long its radius is. All of the points on the circle's circumference are represented by the circle equation.
Given:
$\mathrm{h}=1, \mathrm{k}=2$
Equation of the circle $=(x-1)^{2}+(y-2)^{2}=r^{2} \quad$..(1)
Also, equation (1) passes through $(4,6)$
$\therefore(4-1)^{2}+(6-2)^{2}=r^{2}$
$\Rightarrow 9+16=r^{2}$
$\Rightarrow r=5$
Area of a circle,
$=\pi r^{2}$
$=\pi \times 25$
$=25 \pi$
Therefore, the correct option is (3).
Additional Information: The location of a circle in the Cartesian plane is represented by a circle equation. We can write the equation for a circle if we know the location of the circle's centre and how long its radius is. All of the points on the circle's circumference are represented by the circle equation.
Note: The group of points whose separation from a fixed point has a constant value are represented by a circle. The radius of the circle abbreviated $r$, is a constant that describes this fixed point, which is known as the circle's centre. The formula for a circle$\left(\mathrm{x}-\mathrm{x}_{1}\right)^{2}+\left(\mathrm{y}-\mathrm{y}_{1}\right)^{2}=\mathrm{r}^{2}$ whose centre is at $\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)$
Complete step by step Solution:
Let $(h, k)$ be the centre of a circle with radius $r$.
Thus, its equation will be $(x-h)^{2}+(y-k)^{2}=r^{2}$
The location of a circle in the Cartesian plane is represented by a circle equation. We can write the equation for a circle if we know the location of the circle's centre and how long its radius is. All of the points on the circle's circumference are represented by the circle equation.
Given:
$\mathrm{h}=1, \mathrm{k}=2$
Equation of the circle $=(x-1)^{2}+(y-2)^{2}=r^{2} \quad$..(1)
Also, equation (1) passes through $(4,6)$
$\therefore(4-1)^{2}+(6-2)^{2}=r^{2}$
$\Rightarrow 9+16=r^{2}$
$\Rightarrow r=5$
Area of a circle,
$=\pi r^{2}$
$=\pi \times 25$
$=25 \pi$
Therefore, the correct option is (3).
Additional Information: The location of a circle in the Cartesian plane is represented by a circle equation. We can write the equation for a circle if we know the location of the circle's centre and how long its radius is. All of the points on the circle's circumference are represented by the circle equation.
Note: The group of points whose separation from a fixed point has a constant value are represented by a circle. The radius of the circle abbreviated $r$, is a constant that describes this fixed point, which is known as the circle's centre. The formula for a circle$\left(\mathrm{x}-\mathrm{x}_{1}\right)^{2}+\left(\mathrm{y}-\mathrm{y}_{1}\right)^{2}=\mathrm{r}^{2}$ whose centre is at $\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)$
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