Question

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

Answer 1

Hint: Here we will first find the equation of the circle with given radius and center and then we will find the equation of intersection of the circle with the given line and then put the discriminant of the equation greater than zero to get the distinct and real roots and get the interval of c.
The equation of circle with center (a, b) and radius r is given by:-
\[{\left( {x - a} \right)^2} + {\left( {y - b} \right)^2} = {r^2}\]
Also, the discriminant of general quadratic equation \[a{x^2} + bx + c = 0\] is given by:-
\[D = \sqrt {{b^2} - 4ac} \]

Complete step-by-step answer:
It is given that the circle is centered at \[\left( {2,4} \right)\] and the radius 5.
Now we know that, the equation of circle with center (a, b) and radius r is given by:-
\[{\left( {x - a} \right)^2} + {\left( {y - b} \right)^2} = {r^2}\]
Hence the equation of the given circles is given by:-
\[{\left( {x - 2} \right)^2} + {\left( {y - 4} \right)^2} = {\left( 5 \right)^2}\]
Simplifying it further we get:-
\[{\left( {x - 2} \right)^2} + {\left( {y - 4} \right)^2} = 25\]…………………………………. (1)
Now it is given that the line \[3x - 4y - c = 0\]will meet the circle
Hence, we need to find the intersection of the line and the circle.
Therefore, putting the value of y from the equation of line in equation 1 we get:-
From the equation of line we get:-
\[4y = 3x - c\]
\[ \Rightarrow y = \dfrac{{3x - c}}{4}\]
Putting this value in equation 1 we get:-
\[{\left( {x - 2} \right)^2} + {\left( {\dfrac{{3x - c}}{4} - 4} \right)^2} = 25\]
Taking the LCM we get:-
\[{\left( {x - 2} \right)^2} + {\left( {\dfrac{{3x - c - 16}}{4}} \right)^2} = 25\]
Simplifying it further we get:-
\[{\left( {x - 2} \right)^2} + \dfrac{{{{\left( {3x - c - 16} \right)}^2}}}{{16}} = 25\]
Taking the LCM we get:-
\[16{\left( {x - 2} \right)^2} + {\left( {3x - c - 16} \right)^2} = 25 \times 16\]
\[ \Rightarrow 16{\left( {x - 2} \right)^2} + {\left( {3x - \left( {c + 16} \right)} \right)^2} = 400\]
Now applying the following identity:-
\[{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab\]
We get:-
\[ \Rightarrow 16({x^2} - 4x + 4) + 9{x^2} - 6x(c + 16) + {(c + 16)^2} = 400\]
Solving it further we get:-
\[25{x^2} - (6c + 160)x + {c^2} + 32c - 80 = 0\]
Now we know that for real and distinct roots of an equation the discriminant should be greater than zero and, the discriminant of general quadratic equation \[a{x^2} + bx + c = 0\]is given by:-
\[D = \sqrt {{b^2} - 4ac} \]
Hence, the discriminant of above equation is given by:-
\[{(6c + 160)^2} - 4(25)({c^2} + 32c - 80) > 0\]
Applying the following identity:-
\[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\]
We get:-
\[36{c^2} + 1920c + 25600 - 100{c^2} - 3200c + 8000 > 0\]
Simplifying it further we get:-
\[ \Rightarrow - 64{c^2} - 1280c + 33600 > 0\]
Dividing whole equation by 64 we get:-
\[ - {c^2} - 20c + 525 > 0\]
Now we will multiply the equation by negative sign so the inequality we change.
\[ \Rightarrow {c^2} + 20c - 525 < 0\]
Now solving the equation by middle term split we get:-
\[{c^2} + 35c - 15c - 525 < 0\]
Taking the terms common we get:-
\[c\left( {c + 35} \right) - 15\left( {c + 35} \right) < 0\]
\[ \Rightarrow (c + 35)(c - 15) < 0\]
Solving for c we get:-
\[ \Rightarrow - 35 < c < 15\]
Therefore, proved.

Note: Students should take a note that for a standard equation \[a{x^2} + bx + c = 0\]
If the discriminant is greater than zero \[\left( {D > 0} \right)\] then the equation has real and distinct roots.
If the discriminant is equal to zero \[\left( {D = 0} \right)\] then the equation has real and equal roots.
If the discriminant is less than zero \[\left( {D < 0} \right)\] then the equation has imaginary roots.
Also, students should note that if some quantity in inequality is multiplied by a negative sign then the inequality changes.