Question

Show that the line \[3x - 4y - c = 0\] will meet the circle having a center at \[\left[ {2,4} \right]\] and the radius 5 in real and distinct points if \[ - 35 < c < 15\].

Answer 1

Hint:
We are required to prove the given condition in the question. To do that we will first form the equation of a circle using the given information. Since it is given that the line meets the circle so we will substitute a point from the line into the equation of the circle. Finally, we will use the condition that for real and distinct roots the discriminant is greater than 0. Using this we will prove our given condition.
Formula Used: We will use the following formula to solve our question –
1) Equation of the circle \[{\left[ {x - a} \right]^2} + {\left[ {y - b} \right]^2} = {r^2}\], where \[a\] is the \[x\] coordinate of the center, \[b\] is the\[y\] coordinate of the center and \[r\] is radius of the circle.
2) For a quadratic equation of the kind \[a{x^2} + bx + c = 0\], the discriminant is given by the formula, \[{b^2} - 4ac\], where \[a\] is the coefficient of \[{x^2}\], \[b\] is the coefficient of \[x\] and \[c\] is the constant term.

Complete step by step solution:
We have to prove that the line \[3x - 4y - c = 0\] will meet the circle having a center at \[\left[ {2,4} \right]\] and the radius 5 in real and distinct points only if \[ - 35 < c < 15\].
To do that, we will first form the equation of the given circle.
Here, we are given that, \[\begin{array}{l}a = 2\\b = 4\\c = 5\end{array}\].
Substituting these values in the standard equation of the circle, we get,
\[{\left[ {x - a} \right]^2} + {\left[ {y - b} \right]^2} = {r^2}\]
\[ \Rightarrow {\left[ {x - 2} \right]^2} + {\left[ {y - 4} \right]^2} = {5^2}\]…….\[\left[ 1 \right]\]
As the line intersects or meets the circle, then we will substitute a point from the line into the equation of the circle to get the point of the intersection.
The line given to us is \[3x - 4y - c = 0\].
Finding the value of \[y\] from the above equation of the line, we get,
\[\begin{array}{l}3x - 4y - c = 0\\ \Rightarrow 4y = 3x - c\end{array}\]
\[ \Rightarrow y = \dfrac{{3x - c}}{4}\]…………..\[\left[ 2 \right]\]
Substitute the value of \[y\] from equation \[\left[ 2 \right]\] to equation \[\left[ 1 \right]\]. This will give us the points of intersection,
\[{\left[ {x - 2} \right]^2} + {\left[ {y - 4} \right]^2} = {5^2}\]
\[ \Rightarrow {\left[ {x - 2} \right]^2} + {\left\{ {\left[ {\dfrac{{3x - c}}{4}} \right] - 4} \right\}^2} = 25\]…………….\[\left[ 3 \right]\]
We will solve the equation \[\left[ 3 \right]\] further by taking the LCM of the denominators of the terms within the bracket.
\[\begin{array}{l}{\left[ {x - 2} \right]^2} + {\left\{ {\left[ {\dfrac{{3x - c}}{4}} \right] - 4} \right\}^2} = 25\\ \Rightarrow {\left[ {x - 2} \right]^2} + {\left\{ {\left[ {\dfrac{{3x - c}}{4}} \right] - \dfrac{{4\left[ 4 \right]}}{4}} \right\}^2} = 25\end{array}\]
Now we will take the LCM throughout. On doing so we get,
\[\begin{array}{l} \Rightarrow {\left[ {x - 2} \right]^2} + {\left\{ {\dfrac{{3x - c - 16}}{4}} \right\}^2} = 25\\ \Rightarrow 16{\left[ {x - 2} \right]^2} + {\left\{ {3x - \left[ {c + 16} \right]} \right\}^2} = 25 \times 16\end{array}\]
On squaring the terms and then multiplying them, we get
\[\begin{array}{l} \Rightarrow 16\left[ {{x^2} + 4 - 4x} \right] + \left\{ {9{x^2} + {{\left[ {c + 16} \right]}^2} - 6x\left[ {c + 16} \right]} \right\} = 400\\ \Rightarrow 16{x^2} + 64 - 64x + 9{x^2} + {c^2} + 256 + 32c - 6xc - 96x = 400\end{array}\]
On gathering all the like terms together and then solving them further, we get,
\[ \Rightarrow 16{x^2} + 9{x^2} - 64x - 96x - 6xc + {c^2} + 32c + 64 + 256 - 400 = 0\]
\[ \Rightarrow 25{x^2} - \left[ {160 + 6c} \right]x + {c^2} + 32c - 80 = 0\]………….\[\left[ 4 \right]\]
Now in equation \[\left[ 4 \right]\], we can see that the highest degree is 2, hence this is a quadratic equation.
We know that the roots of the quadratic equation are real and distinct when the discriminant is greater than 0. That is when the discriminant is positive.
Hence, in this question, the intersection points will be real and distinct when discriminant is greater than 0.
We know that the discriminant is given by the formula \[{b^2} - 4ac\].
From equation \[\left[ 4 \right]\], we can infer that,
\[\begin{array}{l}a = 25\\b = 6c + 160\\c = {c^2} + 32c - 80\end{array}\]
Substituting these values in the formula of the discriminant, we get,
\[{\left[ {6c + 160} \right]^2} - 4\left[ {25} \right]\left[ {{c^2} + 32c - 80} \right] > 0\]
On expanding this further by squaring the terms,
\[\begin{array}{l}{\left[ {6c + 160} \right]^2} - 4\left[ {25} \right]\left[ {{c^2} + 32c - 80} \right] > 0\\ \Rightarrow 36{c^2} + 25600 + 1920c - 100\left[ {{c^2} + 32c - 80} \right] > 0\end{array}\]
We will multiply and gather the like terms now. On doing so we get,
\[\begin{array}{l} \Rightarrow 36{c^2} + 25600 + 1920c - 100{c^2} - 3200c + 8000 > 0\\ \Rightarrow 36{c^2} - 100{c^2} + 1920c - 3200c + 25600 + 8000 > 0\end{array}\]
On solving the terms further and taking out 64 common from all the terms, we get,
\[\begin{array}{l} \Rightarrow - 64{c^2} - 1280c + 33600 > 0\\ \Rightarrow 64\left[ { - {c^2} - 20c + 525} \right] > 0\end{array}\]
On dividing both sides of the equation with 64, we get,
\[\begin{array}{l} \Rightarrow - {c^2} - 20c + 525 > 0\\ \Rightarrow {c^2} + 20c + 525 < 0\end{array}\]
Now we will factorize the above equation using the mid-term splitting concept.
\[\begin{array}{l}{c^2} + 20c + 525 < 0\\ \Rightarrow {c^2} + \left[ {35 - 15} \right]c + 525 < 0\\ \Rightarrow {c^2} + 35c - 15c + 525 < 0\end{array}\]
On resolving the terms as factors, we get,
\[ \Rightarrow \left[ {c + 35} \right]\left[ {c - 15} \right] < 0\]
Using the zero product rule, we can write
\[\begin{array}{l}c + 35 < 0\\c > - 35\end{array}\]
or
\[\begin{array}{c}c - 15 < 0\\c < 15\end{array}\]

Hence, proved that our line will intersect the circle in real and distinct points if \[ - 35 < c < 15\].

Note:
Whenever we have to find the point of intersection of any two geometric figures, be they lines, or circle or any other form, always remember that we should calculate the value of the variable from one equation and then put it into the second one. Upon solving, our answer will be your required point of intersection. Also, we must take care of negative signs when finding the value of \[c\]. Any change in sign can change the range of \[c\].