Question

Choose the correct option from the below given options by solving the following question:
If the line \[3x - 4y - k = 0\]\[,\left( {k > 0} \right)\] touches the circle \[{x^2} + {y^2} - 4x - 8y - 5 = 0\] at \[\left( {a,b} \right)\], then \[k + a + b\] equal to?
A. \[20\]
B. \[22\]
C. \[ - 30\]
D. \[ - 28\]

Answer 1

Hint: Convert the given equation of the circle to the standard equation of the circle form and find the centre and radius of the circle. Then, substitute the centre of the circle points in the equation of the line given to find the length. Consider a parameter by comparing the obtained equation to the equation given above. After finding the value of the parameter, substitute them in the point given to find the value of the coordinates of the point. Take the given question, substitute the points in the expression to get the final answer.

Complete step-by-step solution:
Given equations;
The equation of the line- \[3x - 4y - k = 0\]
The equation of the circle- \[{x^2} + {y^2} - 4x - 8y - 5 = 0\]
Given point:
\[\left( {a,b} \right)\]
Let us calculate the details from the circle. From the equation of the circle,
We can write the equation of the circle as:
\[\left( {{x^2} - 2\left( x \right)\left( 2 \right) + 4} \right) + \left( {{y^2} - 2\left( y \right)\left( 4 \right) + 16} \right) - 4 - 16 - 5 = 0\]
Adding and subtracting \[4\]and \[16\] to the respective \[x,y\]terms respectively to get the standard equation of the circle.
Now, simplifying it, we can write the equation as:
\[{\left( {x - 2} \right)^2} + {\left( {y - 4} \right)^2} - 25 = 0\]
Bringing the constant value to the right-hand side, we get;
\[{\left( {x - 2} \right)^2} + {\left( {y - 4} \right)^2} = 25\]
\[25\]can be written as \[{5^2}\]
\[ \Rightarrow {\left( {x - 2} \right)^2} + {\left( {y - 4} \right)^2} = {5^2}\]
We get the radius \[r = 5\]
\[h = 2\]
\[k = 4\]
So, we have the centre of the circle as \[\left( {h,k} \right) = \left( {2,4} \right)\]
It is given that the line touches the circle, that means it becomes the radius. Substituting the centre and finding the distance of the perpendicular from the centre, which is also a radius, we get;
\[ \pm S = \dfrac{{3 \times 2 - 4 \times 4 - k}}{{\sqrt {{3^2} + {4^2}} }}\]
Substituting the radius and simplifying the equation, we get;
$\Rightarrow$\[ \pm 5 = \dfrac{{6 - 16 - k}}{{\sqrt {9 + 16} }}\]
Now, simplifying the square root in the equation, equation, we get;
$\Rightarrow$\[ \pm 5 = \dfrac{{ - 10 - k}}{{\sqrt {25} }}\]
Taking the square root out, we get;
$\Rightarrow$\[ \pm 5 = \dfrac{{ - 10 - k}}{5}\]
Taking the denominator to the left-hand side, we get;
$\Rightarrow$\[ - 10 - k = \pm 25\]
Now, taking the variables to one side and the constants to other side, we get;
$\Rightarrow$\[ - k = \pm 25 + 10\]
Taking \[ + 25\], we get;
$\Rightarrow$\[k = - 35\]
Taking \[ - 25\], we get;
$\Rightarrow$\[k = 15\]
We take \[k = 15\], since it is mentioned that \[k > 0\] and \[ - 35\]\[ < 0\]. Therefore, \[k = 15\] is considered.
Now, if we consider a point \[\left( {a,b} \right)\], the equation of the tangent at \[\left( {a,b} \right)\] to the circle, we can give it by substituting the point \[\left( {a,b} \right)\] in the equation of the circle.
\[ \Rightarrow xa + yb - 2(x + a) - 4(y + b) - 5 = 0\]
Expanding the equation, we get;
\[ \Rightarrow ax + by - 2x - 2a - 4y - 4b - 5 = 0\]
Taking the common terms of all the variables out and grouping the terms, we get;
\[ \Rightarrow (a - 2)x + (b - 4)y - (2a + 4b + 5) = 0\]
Now, this is compared with the given line \[3x - 4y - k = 0\].
Let us compare and then take a parameter \[l\].
\[ \Rightarrow \dfrac{{a - 2}}{3} = \dfrac{{b - 4}}{{ - 4}} = \dfrac{{2a + 4b + 5}}{k} = l\]
Equating each term to the parameter, we have;
\[ \Rightarrow \dfrac{{a - 2}}{3}\]\[ = l\]
\[ \Rightarrow a = 3l + 2\]
Next equation, we have;
\[ \Rightarrow \dfrac{{b - 4}}{{ - 4}}\]\[ = l\]
\[ \Rightarrow b = - 4l + 4\]
The last equation, we have;
\[ \Rightarrow \dfrac{{2a + 4b + 5}}{k} = l\]
\[ \Rightarrow 2a + 4b + 5 = kl\]
Substituting \[k\], we get;
\[ \Rightarrow 2a + 4b = 15l - 5\]
Substituting the values of \[a,b\] in the above equation, we get;
\[ \Rightarrow 2\left( {3l + 2} \right) + 4\left( { - 4l + 4} \right) = 15l - 5\]
Simplifying the equation, we get;
\[ \Rightarrow 6l + 4 - 16l + 16 = 15l - 5\]
Bringing the variables to one side and the constants to the other side, we get;
\[ \Rightarrow 6l - 16l - 15l = - 5 - 16 - 4\]
Simplifying the equation, we get;
\[ \Rightarrow \]\[ - 25l = - 25\]
\[ \Rightarrow \]\[l = 1\]
Now, we have, \[l = 1\]. From this, we can find the values of \[a\]and \[b\]
\[ \Rightarrow a = 3l + 2\]
\[ \Rightarrow a = 3 + 2 = 5\]
\[ \Rightarrow b = - 4l + 4\]
\[ \Rightarrow b = - 4 + 4 = 0\]
We have the values of \[a = 5,b = 0,k = 15\]
We have to find \[k + a + b\]. Substituting the values of \[k,a,b\], we get;
\[15 + 5 + 0 = 20\]
Therefore, \[k + a + b = 20\].

The correct option is A.

Note: We have the standard equation of the circle as \[{\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}\], where, \[\left( {h,k} \right)\] are the coordinates of the centre of the circle and \[r\] is the radius of the circle. We compare the given equation of the circle to the standard equation of the circle, if in case they are not in the same format, we have to convert the given equation into the standard equation.