Question

If the area of the triangle formed by the positive x-axis, the normal and the tangent to the circle ${(x-2)}^2+{(y-3)}^2=25$ at the point $(5,7)$ is $A$, then $24A$ is equal to

Answer 1

Step 1: Draw a schematic diagram diagram with the given date in the question

From the given equation of circle ${(x-2)}^2+{(y-3)}^2=25$, the centre is at $\left(2,3\right)$ and its radius is $5\text{units}$, then

Step 2: Find the equation of normal,

From point $\left(2,3\right)$ and $(5,7)$, the equation of normal is

$\begin{array}{l}y-3=\frac{7-3}{5-2}(x-2)\\ y-3=\frac{4}{3}(x-2)\mathrm{........}\left(i\right)\end{array}$

When $y=0$, then

$\begin{array}{c}-9=4x-8\\ -9+8=4x\\ x=-\frac{1}{4}\end{array}$

Then the point of intersection with the x-axis, that is point $Q\left(-\frac{1}{4},0\right)$

Step 3: Find the equation of tangent,

we know that the product of slopes of normal and tangent is equal to $-1$, as both are perpendicular to each other.

From equation (i) the slope of normal, $m_1=\frac{4}{3}$,

we know $m_1\times m_2=-1$, then $m_2=-\frac{3}{4}$,

Using the slope $m_2=-\frac{3}{4}$ and a point $(5,7)$, the equation of tangent can be written as

$y-7=-\frac{3}{4}(x-5).....\left(ii\right)$

When $y=0$, then

$\begin{array}{c}28=3x-15\\ 43=3x\\ x=\frac{43}{3}\end{array}$

Then the point of intersection with the x-axis, that is point $R\left(\frac{43}{3},0\right)$

Step 4: Calculate the Area PQR,

Area of triangle $=\frac{1}{2}\times base\times height$

Using the figure, Height $=7$, and

$\begin{array}{rcl}\mathrm{base}& =& \frac{43}{3}-\left(-\frac{1}{4}\right)\\ & =& \frac{172+3}{12}\\ & =& \frac{175}{12}\end{array}$

Therefore, the area $A=\frac{1}{2}\times \frac{175}{12}\times 7$, then

$\begin{array}{rcl}24A& =& 24\times \frac{1}{2}\times \frac{175}{12}\times 7\\ & =& 175\times 7\\ & =& 1225\end{array}$

Hence, $24A=1225\text{square unit}$.