VerifiedHint: In this question first draw the pictorial representation of the given problem which will help us to find out what we have to find out then compare the given equation of circle with the standard equation of circle which is given as ${x^2} + {y^2} + 2gx + 2fy + a = 0$where (-g, -f) is the center of the circle and $\sqrt {{g^2} + {f^2} - a} $ is the radius of the circle, later on use the perpendicular distance formula from a given point on the perpendicular line.
Complete step-by-step answer:
A given point on the perpendicular line.
Given equation of circle is
${x^2} + {y^2} - 2cx - 2cy + {c^2} = 0$
The pictorial representation of the given problem is shown above,
As the line is touching the circle so it is the tangent of the circle and makes a right angle with the radius of the circle as shown in the figure.
Let the line touch the circle at the point A.
Now as we know that the general equation of the circle is
${x^2} + {y^2} + 2gx + 2fy + a = 0$
Where (-g, -f) is the center of the circle and radius r = $\sqrt {{g^2} + {f^2} - a} $
So on comparing we have,
g = - c, f = - c and a = c2.
So the center of the circle is O = (c, c) as shown in the figure and the radius is
$\sqrt {{{\left( { - c} \right)}^2} + {{\left( { - c} \right)}^2} - {c^2}} = \sqrt {{c^2} + {c^2} - {c^2}} = \sqrt {{c^2}} = c$
Now it is also given that the center lying in the first quadrant as shown in figure therefore c must be positive value.
So the radius of the circle is also c.
Now the equation of given line is
$\dfrac{x}{3} + \dfrac{y}{4} = 1$
Now this equation is written as
$4x + 3y - 12 = 0$
Now the perpendicular distance which is nothing but radius from the center of the circle on the line is calculated as,
$c = \dfrac{{\left| {a{x_1} + b{y_1} + p} \right|}}{{\sqrt {{a^2} + {b^2}} }}$, (where (a) and (b) are the coefficient of x and y respectively, p is the constant of the equation and x1 and y1 are the center of the circle).
So substitute the values we have,
$ \Rightarrow c = \dfrac{{\left| {4c + 3c - 12} \right|}}{{\sqrt {{4^2} + {3^2}} }} = \dfrac{{\left| {4c + 3c - 12} \right|}}{{\sqrt {25} }} = \dfrac{{\left| {4c + 3c - 12} \right|}}{5}$
$ \Rightarrow 5c = \left| {7c - 12} \right|$
Now when modulus opens as positive
$ \Rightarrow 5c = + \left( {7c - 12} \right)$
$ \Rightarrow 7c - 5c = 12$
$ \Rightarrow 2c = 12$
$ \Rightarrow c = 6$
Now when modulus opens as negative
$ \Rightarrow 5c = - \left( {7c - 12} \right)$
$ \Rightarrow 7c + 5c = 12$
$ \Rightarrow 12c = 12$
$ \Rightarrow c = 1$
So the required value of c is either 1 or 6.
So this is the required answer.
Hence option (1) and (4) are correct.
Note – Whenever we face such types of questions always recall the general equation of the circle than compare it with the given circle equation and calculate its center and radius of the circle as above the use the concept if a line touches the circle then the line makes a right angle triangle with the radius of the circle so use the perpendicular distance formula between center and given line $d = \dfrac{{\left| {a{x_1} + b{y_1} + p} \right|}}{{\sqrt {{a^2} + {b^2}} }}$ and simplify, we will get the required answer.
