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Another problem on circles

DrunkenOldFool

New member
Feb 6, 2012
20
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

Active member
Jan 27, 2012
95
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 $(-8,0)$.
 
Last edited:

DrunkenOldFool

New member
Feb 6, 2012
20
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 $(-8,0)$.
Thank You!!!!