Question

Find the number of integral values of \[k\], for which, the line \[3x + 4y = k\] intersects the circle \[{x^2} + {y^2} - 2x - 4y + 4 = 0\] at two distinct points.

Answer 1

Hint: First the center and the radius of the circle are found from the given data and then the distance of the center in terms of variable \[k\] is calculated using the formulae and inequality is established with the radius of the circle to find the number of integral values of \[k\] so that the line will intersect the circle at two distinct points.

Formula Used:The distance \[d\] of a point \[({x_1},{y_1})\] from a line \[ax + by + c = 0\] is calculated by using the formulae \[d = \dfrac{{\left| {a{x_1} + b{y_1} + c} \right|}}{{\sqrt {{a^2} + {b^2}} }}\]. The generalized equation of a circle is \[{(x - h)^2} + {(y - k)^2} = {r^2}\], where, \[(h,k)\]and \[r\] are the center and radius of the circle respectively.

Complete step-by-step solution:
We have been given the equation of the circle as \[{x^2} + {y^2} - 2x - 4y + 4 = 0\].
First, we will find the center and the radius of the circle from the given equation by converting it into the generalized equation \[{(x - h)^2} + {(y - k)^2} = {r^2}\], where, \[(h,k)\]and \[r\]are the center and radius of the circle respectively.
\[{x^2} + {y^2} - 2x - 4y + 4 = 0\]
\[ \Rightarrow ({x^2} - 2x + 1) + ({y^2} - 4y + 4) - 5 + 4 = 0\]
\[ \Rightarrow {(x - 1)^2} + {(y - 2)^2} = {1^2}\]
Comparing the above equation with the generalized equation of the circle, we have
\[h = 1\], \[k = 2\] and \[r = 1\]
So, the center of the circle is \[(1,2)\] and its radius is equal to \[1\].
Now, we will calculate the distance of the center \[(1,2)\] from the given line \[3x + 4y = k\] by using the formulae.
Let, \[d = \] the distance of the center from the line
Assigning the values to the variables of the formulae,
\[{x_1} = 1\], \[{y_1} = 2\], \[r = 1\], \[a = 3\], \[b = 4\] and \[c = - k\]
Substituting the values in the formulae, we have
\[d = \dfrac{{\left| {(3 \times 1) + (4 \times 2) + ( - k)} \right|}}{{\sqrt {{3^2} + {4^2}} }}\]
\[ \Rightarrow d = \dfrac{{\left| {3 + 8 - k} \right|}}{{\sqrt {{5^2}} }}\]
\[ \Rightarrow d = \dfrac{{\left| {11 - k} \right|}}{5}\]
We know that the line \[3x + 4y = k\] will intersect the circle at two distinct points if and only if \[d\] is less than the radius of the circle.
So, \[d < 1\] (Since \[d = \dfrac{{\left| {11 - k} \right|}}{5}\])
\[ \Rightarrow \dfrac{{\left| {11 - k} \right|}}{5} < 1\]
\[ \Rightarrow \left| {11 - k} \right| < 5\]
\[ \Rightarrow - 5 < k - 11 < 5\]
Further solving the above, we have
\[ \Rightarrow - 5 + 11 < k < 5 + 11\]
\[ \Rightarrow 6 < k < 16\]
\[ \Rightarrow k = 7,8,9,10,11,12,13,14,15\] (total nine in number)

Hence, the total number of integral values of \[k\] is nine.

Note: If the distance of the center of a circle from a line is less than its radius, then the line will definitely intersect the circle at two distinct points. But, if this distance is equal to the radius of the circle, then the line will only touch the circle at one point instead of intersecting it at two distinct points.