Question

The equation of a circle and a line are ${{\text{x}}^2} + {{\text{y}}^2} - 8{\text{x + 2y + 12 = 0 }}$ and x-2y-1=0. Determine whether the line is a chord or a tangent or does not meet the circle at all.

Answer 1

Hint: First we find the centre of the given circle and then find the distance of point from the line x-2y-1 =0. If the distance comes out to be equal to the radius of the circle then the given line is tangent to the circle.

Complete step-by-step answer:
The general equation of a circle is given as:
${{\text{x}}^2} + {{\text{y}}^2}{\text{ + 2gx + 2fy + c = 0 }}$ (1)
The centre of this circle is given as: O (-g ,-f)
The given equation of circle is ${{\text{x}}^2} + {{\text{y}}^2} - 8{\text{x + 2y + 12 = 0 }}$
Comparing this equation with equation 1, we get:
2g = 8
$\Rightarrow$ g = 4 and
2f = 2
$\Rightarrow$f =1
$\therefore$ Centre O (-4 , -1)
We know that length of perpendicular drawn from point(${{\text{x}}_1},{{\text{y}}_1}$) is given by:
d = $\dfrac{{{\text{|a}}{{\text{x}}_1} + {\text{b}}{{\text{y}}_1} + c|}}{{\sqrt {{{\text{a}}^2} + {{\text{b}}^2}} }}$
Putting the values in above equation, we get:
d = $\dfrac{{{\text{| - 4}} + ( - 2 \times - 4) - 1|}}{{\sqrt {{1^2} + {2^2}} }} = \sqrt 5$
Now, we calculate the radius of the circle.
r= $\sqrt {{{\text{g}}^2} + {{\text{f}}^2} - c} = \sqrt {{4^2} + {1^2} - 12} = \sqrt 5$

$\because$ The distance from centre of circle to the given line is equal to the radius of the radius.
So, the given line is tangent to the given circle.

Note- In such a type of question, you should know how to check for tangency of a line to a circle. For a circle ${{\text{x}}^2} + {{\text{y}}^2} = {r^2}$ and a line y = mx +c, the condition of tangency is given as:
${{\text{c}}^2} = {r^2}(1 + {m^2})$ . And if general equation of the circle is given i.e${{\text{x}}^2} + {{\text{y}}^2}{\text{ + 2gx + 2fy + c = 0 }}$then in this case calculate distance from (-g,-f) to the line and it must be equal to radius of circle.