Question

Find the difference of the maximum and the minimum distance of a point on the line $x+y-1=0$, where it cuts x-axis with respects to the circle ${{x}^{2}}+{{y}^{2}}-6x-8y+24=0$?
(a) 1,
(b) 2,
(c) $2\sqrt{5}$,
(d) $3\sqrt{5}$.

Answer 1

Hint: We start solving with the drawing all the information to get a better view about it. We then find the point of intersection of line $x+y-1=0$, where it cuts x-axis. We then find the center and radius of the circle ${{x}^{2}}+{{y}^{2}}-6x-8y+24=0$. We use the fact that the difference between the maximum and minimum distance is equal to the diameter of the circle to get the required result.

Complete step-by-step answer:
According to the problem, we have to find the difference of the maximum and minimum distance of a point on the line $x+y-1=0$ where it cuts x-axis with respects to the circle ${{x}^{2}}+{{y}^{2}}-6x-8y+24=0$.
Let us draw the given information to get a better view.

Let us find the point of intersection of line $x+y-1=0$ where it cuts x-axis. We know that on x-axis the value of y is 0. Let us substitute this in $x+y-1=0$.
$\Rightarrow x+0-1=0$.
$\Rightarrow x-1=0$.
$\Rightarrow x=1$.
The point where the line $x+y-1=0$ cuts x-axis is $B\left( 1,0 \right)$ ---(1).
Let us find the center and radius of the circle ${{x}^{2}}+{{y}^{2}}-6x-8y+24=0$.
$\Rightarrow {{x}^{2}}-6x+9+{{y}^{2}}-8y+16-1=0$.
$\Rightarrow {{\left( x-3 \right)}^{2}}+{{\left( y-4 \right)}^{2}}=1$.
$\Rightarrow {{\left( x-3 \right)}^{2}}+{{\left( y-4 \right)}^{2}}={{1}^{2}}$.
We know that if the equation of the circle is ${{\left( x-a \right)}^{2}}+{{\left( y-b \right)}^{2}}={{r}^{2}}$, then the center of the circle is $\left( a,b \right)$ and the radius of the circle is r units.
So, we have center and radius of the circle ${{x}^{2}}+{{y}^{2}}-6x-8y+24=0$ as $A\left( 3,4 \right)$ and 1 unit.
We can see from the figure that the minimum distance of the point from the $B\left( 1,0 \right)$ to the circle ${{x}^{2}}+{{y}^{2}}-6x-8y+24=0$ is the parallel distance between the line $x+y-1=0$ and the tangent parallel to this circle below the center.
We can see from the figure that the maximum distance of the point from the $B\left( 1,0 \right)$ to the circle ${{x}^{2}}+{{y}^{2}}-6x-8y+24=0$ is the parallel distance between the line $x+y-1=0$ and the tangent parallel to this circle above the center.
We need the difference between the maximum and minimum distances which will be the distance between these tangents. We know that the distance between parallel tangents is equal to the diameter of the circle.
So, we have $d=2\times radius$.
$\Rightarrow d=2\times 1$.
$\Rightarrow d=2$ units.
We have found the difference of the maximum and the minimum distance of a point on the line $x+y-1=0$, where it cuts x-axis with respects to the circle ${{x}^{2}}+{{y}^{2}}-6x-8y+24=0$ is 2 units.
The correct option for the given problem is (b).

Note:We can also solve the distance alternatively as follows.
Let us find the distance between A and B.
$AB=\sqrt{{{\left( 3-1 \right)}^{2}}+{{\left( 4-0 \right)}^{2}}}$.
$AB=\sqrt{{{2}^{2}}+{{4}^{2}}}$.
$AB=\sqrt{4+16}$.
$AB=\sqrt{20}$.
$AB=\sqrt{4\times 5}$.
$AB=2\sqrt{5}$.
We know that the minimum and maximum distances of the point with respect to the circle is $d-r$ and $d+r$, where d is the distance between the center of the circle and the point.
So, we have minimum and maximum distances as $2\sqrt{5}-1$ and $2\sqrt{5}+1$.
Let us find the difference of $2\sqrt{5}-1$ and $2\sqrt{5}+1$.
We have $2\sqrt{5}+1-\left( 2\sqrt{5}-1 \right)$.
$\Rightarrow 2\sqrt{5}+1-2\sqrt{5}+1$.
$\Rightarrow 2$ units.
So, we have found the difference as 2 units.