Solveeit Logo
Login

Question

Question: If the line \( 3x - 4y - k = 0 \) , \( \left( {k > 0} \right) \) touches the circle \( {x^2} + {y^2}...

If the line 3x−4y−k=03x - 4y - k = 0 , (k>0)\left( {k > 0} \right) touches the circle x2+y2−4x−8y−5=0{x^2} + {y^2} - 4x - 8y - 5 = 0 at (a,b),\left( {a,b} \right), then k+a+bk + a + b is equal to
A. 2020
B. 2222
C. −30- 30
D. −28- 28

Explanation

Solution

Hint : In the question, we are given that a line is touching a circle at a certain point. Now, that line is the tangent of that circle. So, using the formula for the tangent of the circle. And then comparing it with the given equation can solve the question easily.

Formula used:
General equation of circle x2+y2+2gx+2fy+c=0{x^2} + {y^2} + 2gx + 2fy + c = 0
Equation of tangent which is touching the circle at the point (x1,y1)\left( {{x_1},{y_1}} \right) is xx1+yy1+g(x+x1)+f(y+y1)+c=0x{x_1} + y{y_1} + g(x + {x_1}) + f(y + {y_1}) + c = 0

Complete step-by-step answer :
Firstly, reviving the concepts for the circle. The general equation of the circle is
x2+y2+2gx+2fy+c=0{x^2} + {y^2} + 2gx + 2fy + c = 0 …(1)\ldots \left( 1 \right) where (−g,−f)\left( { - g, - f} \right) is the center and radius is given by R=g2+f2−cR = \sqrt {{g^2} + {f^2} - c} and g, f, c are the constants
In the question, we are provided an equation of circle x2+y2−4x−8y−5=0{x^2} + {y^2} - 4x - 8y - 5 = 0 Comparing the given equation with the general equation (1)\left( 1 \right) .
g=−2g = - 2 and f=−4f = - 4
According to the question, we are given that the line 3x−4y−k=03x - 4y - k = 0 touches the circle at (a,b),\left( {a,b} \right), so now using the general equation for the tangent which is touching the given circle is
ax+by−2(x+a)−4(y+b)−5=0ax + by - 2(x + a) - 4(y + b) - 5 = 0
Solving to form a linear equation of line,
(a−2)x+(b−4)y+(−2a−4b−5)=0(a - 2)x + (b - 4)y + ( - 2a - 4b - 5) = 0
Comparing this equation with the given line 3x−4y−k=03x - 4y - k = 0
Coefficient of x,
a−2=3 a=3+2=5   a - 2 = 3 \\\ a = 3 + 2 = 5 \;
Coefficient of y,
b−4=−4 b=−4+4=0   b - 4 = - 4 \\\ b = - 4 + 4 = 0 \;
Constant terms,
k=2a+4b+5k = 2a + 4b + 5
Using the above calculated values of aa and bb
k=2(5)+4(0)+5 =10+5=15   k = 2(5) + 4(0) + 5 \\\ = 10 + 5 = 15 \;
But in the question, we are asked to calculate the value of k+a+bk + a + b
So, substituting the values
15+5+0=2015 + 5 + 0 = 20
So, the correct option is A.
So, the correct answer is “Option A”.

Note : The comparison of the equations with the general equation should be done properly. Because mostly students left the signs behind them which led to direct wrong solutions. While solving mathematical questions, the calculations should be done with full attention.