Question

If \[x\cos \alpha + y\sin \alpha = p\]is a tangent to a circle \[{x^2} + {y^2} = 2q(x\cos \alpha + y\sin \alpha )\] then the set of possible values of p is

Answer 1

Hint: In a quadratic equation\[a{x^2} + bx + c = 0\], its discriminant is \[\sqrt {{b^2} - 4ac} \]. In a second-degree quadratic equation, we know it has two values for its unknown value. When the discriminant value is zero, then the circle and parabola are tangent with each other. To solve this problem, we need to understand the relation between circles and lines, how this can be explained using a quadratic equation.

Complete step by step answer:
Now we look at the given question, we are given an equation \[x\cos \alpha + y\sin \alpha = p\].
Let’s find the value of ‘y’ from this equation.
From this we have \[y = \dfrac{{p - x\cos \alpha }}{{\sin \alpha }}\]
Now put this value of ‘p’ in the given equation of the circle.
That is \[{x^2} + {y^2} = 2q(x\cos \alpha + y\sin \alpha )\]becomes\[{x^2} + {\left( {\dfrac{{p - x\cos \alpha }}{{\sin \alpha }}} \right)^2} = 2qx\cos \alpha + 2q\left( {\dfrac{{p - x\cos \alpha }}{{\sin \alpha }}} \right)\sin \alpha \]
That is \[{x^2} + \left( {\dfrac{{{p^2} - 2px\cos \alpha + {x^2}{{\cos }^2}\alpha }}{{{{\sin }^2}\alpha }}} \right) - 2qx\cos \alpha - 2qp + 2qx\cos \alpha = 0\]
Or we will get \[{x^2}{\sin ^2}\alpha + {p^2} - 2px\cos \alpha + {x^2}{\cos ^2}\alpha - 2qp{\sin ^2}\alpha = 0\]
\[{x^2}\left( {{{\sin }^2}\alpha + {{\cos }^2}\alpha } \right) + {p^2} - 2px\cos \alpha - 2qp{\sin ^2}\alpha = 0\]
\[{x^2} - 2px\cos \alpha + {p^2} - 2qp{\sin ^2}\alpha = 0\]
This is of the form \[a{x^2} + bx + c = 0\]where$a\; = 1,\;b\; = - 2p\cos \propto ,c = {p^2} - 2qp{\sin ^2} \propto $
The equation of discriminant is \[\sqrt {{b^2} - 4ac} \]
That is ${\left( { - 2p{\text{cos}}\alpha } \right)^2} - 4 \times 1 \times \left( {{p^2} - 2qp{\text{si}}{{\text{n}}^2}\alpha } \right)$
For being tangent, the value of the discriminant should be zero. Therefore, we have to equate this equation to zero.
That is \[{\left( { - 2p\cos \alpha } \right)^2} - 4 \times 1 \times \left( {{p^2} - 2qp{{\sin }^2}\alpha } \right) = 0\]
\[4{p^2}{\cos ^2}\alpha - 4{p^2} + 8qp{\sin ^2}\alpha = 0\]
\[2qp{\sin ^2}\alpha - {p^2}(1 - {\cos ^2}\alpha ) = 0\]
But we know \[1 - {\cos ^2}\alpha = {\sin ^2}\alpha \]
Then the equation will become \[2qp{\sin ^2}\alpha - {p^2}{\sin ^2}\alpha = 0\]
Or \[{\sin ^2}\alpha (2qp - {p^2}) = 0\]
But \[{\sin ^2}\alpha \ne 0\]
Therefore \[2qp - {p^2} = 0\]
That is \[p\left( {2q - p} \right) = 0\]
On solving we get values of p as ‘\[p{\text{ }} = {\text{ }}0\]’ and ‘\[p{\text{ }} = {\text{ }}2q\]’

Note:
When the value of the discriminant is positive, we have two solutions for the equation. If the discriminant value is negative then there will be no solution and when the discriminant value is zero, then there is only one solution. Also, the equation of a circle with center (h, k) and radius ‘r’ is \[{\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}\].