- #1
DrunkenOldFool
- 20
- 0
Tangents are drawn to the circle $x^2+y^2=32$ from a point $A$ lying on the x-axis. The tangents cut the y-axis at a point $B$ and $C$, then find the coordinate(s) of $A$ such that the area of $\Delta ABC$ is minimum.
sbhatnagar said:The point $A$ lies on the x-axis so we can assume it to be $(k,0)$. The next step is to find the equation of tangents to the circle $x^2+y^2=32$ passing through $(k,0)$. Note that $y=m(x-k)$ is any general line passing through $(k,0)$. The perpendicular distance from the center of the circle to the line $y=m(x-k)$ will be equal to $\displaystyle \Bigg| \dfrac{mk}{\sqrt{1+m^2}}\Bigg|$. If this distance is equal to the radius of the circle then it will be a tangent.
\[ \displaystyle \Bigg| \dfrac{mk}{\sqrt{1+m^2}}\Bigg|= 4\sqrt{2}\]
From here you will get $\displaystyle m=\pm \frac{4\sqrt{2}}{\sqrt{k^2-32}}$. So the two tangents are $y\sqrt{k^2-32}=\pm 4\sqrt{2}(x-k)$. The points $B$ and $C$ will come out to be $\left(0, \dfrac{4\sqrt{2}k}{\sqrt{k^2-32}}\right)$ and $\left(0, \dfrac{-4\sqrt{2}k}{\sqrt{k^2-32}}\right)$.
The area of $\Delta ABC$ will be equal to
\[ a(k)=\frac{1}{2} BC \times AO=\frac{1}{2}\frac{8\sqrt{2}k}{\sqrt{k^2-32}}k=\frac{4\sqrt{2}k^2}{\sqrt{k^2-32}}\]
We seek the value of $k$ for which $a(k)$ is minimum. Calculate the derivative of $a(k)$ and then set it to 0.
\[ 0= \frac{8k\sqrt{2(k^2-32)} -\frac{4\sqrt{2}k^3}{\sqrt{k^2-32}}}{k^2-32}\]
From here you should get $k=\pm 8$. The minimum area is obtained at $(8,0)$ and $(-8,0)$.
The formula for finding the area of a circle is A = πr², where A represents the area and r represents the radius of the circle.
To calculate the circumference of a circle, you can use the formula C = 2πr, where C represents the circumference and r represents the radius of the circle.
If you know the circumference of a circle, you can find the radius by dividing the circumference by 2π. So, the formula would be r = C / 2π.
The diameter of a circle is twice the length of the radius. In other words, the diameter is equal to 2r. This relationship is important in many circle calculations.
Circles are used in many real-world applications, such as calculating the area of a garden, determining the circumference of a tire, or finding the volume of a cylinder. Understanding the properties and formulas of circles can help you solve a variety of problems in different fields of science and engineering.