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Question: If \[x\cos \alpha + y\sin \alpha = p\]is a tangent to a circle \[{x^2} + {y^2} = 2q(x\cos \alpha + y...

If xcos⁡α+ysin⁡α=px\cos \alpha + y\sin \alpha = pis a tangent to a circle x2+y2=2q(xcos⁡α+ysin⁡α){x^2} + {y^2} = 2q(x\cos \alpha + y\sin \alpha ) then the set of possible values of p is

Explanation

Solution

In a quadratic equationax2+bx+c=0a{x^2} + bx + c = 0, its discriminant is b2−4ac\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 xcos⁡α+ysin⁡α=px\cos \alpha + y\sin \alpha = p.
Let’s find the value of ‘y’ from this equation.
From this we have y=p−xcos⁡αsin⁡αy = \dfrac{{p - x\cos \alpha }}{{\sin \alpha }}
Now put this value of ‘p’ in the given equation of the circle.
That is x2+y2=2q(xcos⁡α+ysin⁡α){x^2} + {y^2} = 2q(x\cos \alpha + y\sin \alpha )becomesx2+(p−xcos⁡αsin⁡α)2=2qxcos⁡α+2q(p−xcos⁡αsin⁡α)sin⁡α{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 x2+(p2−2pxcos⁡α+x2cos⁡2αsin⁡2α)−2qxcos⁡α−2qp+2qxcos⁡α=0{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 x2sin⁡2α+p2−2pxcos⁡α+x2cos⁡2α−2qpsin⁡2α=0{x^2}{\sin ^2}\alpha + {p^2} - 2px\cos \alpha + {x^2}{\cos ^2}\alpha - 2qp{\sin ^2}\alpha = 0
x2(sin⁡2α+cos⁡2α)+p2−2pxcos⁡α−2qpsin⁡2α=0{x^2}\left( {{{\sin }^2}\alpha + {{\cos }^2}\alpha } \right) + {p^2} - 2px\cos \alpha - 2qp{\sin ^2}\alpha = 0
x2−2pxcos⁡α+p2−2qpsin⁡2α=0{x^2} - 2px\cos \alpha + {p^2} - 2qp{\sin ^2}\alpha = 0
This is of the form ax2+bx+c=0a{x^2} + bx + c = 0wherea  =1,  b  =−2pcos⁡∝,c=p2−2qpsin⁡2∝a\; = 1,\;b\; = - 2p\cos \propto ,c = {p^2} - 2qp{\sin ^2} \propto
The equation of discriminant is b2−4ac\sqrt {{b^2} - 4ac}
That is (−2pcosα)2−4×1×(p2−2qpsin2α){\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 (−2pcos⁡α)2−4×1×(p2−2qpsin⁡2α)=0{\left( { - 2p\cos \alpha } \right)^2} - 4 \times 1 \times \left( {{p^2} - 2qp{{\sin }^2}\alpha } \right) = 0
4p2cos⁡2α−4p2+8qpsin⁡2α=04{p^2}{\cos ^2}\alpha - 4{p^2} + 8qp{\sin ^2}\alpha = 0
2qpsin⁡2α−p2(1−cos⁡2α)=02qp{\sin ^2}\alpha - {p^2}(1 - {\cos ^2}\alpha ) = 0
But we know 1−cos⁡2α=sin⁡2α1 - {\cos ^2}\alpha = {\sin ^2}\alpha
Then the equation will become 2qpsin⁡2α−p2sin⁡2α=02qp{\sin ^2}\alpha - {p^2}{\sin ^2}\alpha = 0
Or sin⁡2α(2qp−p2)=0{\sin ^2}\alpha (2qp - {p^2}) = 0
But sin⁡2α≠0{\sin ^2}\alpha \ne 0
Therefore 2qp−p2=02qp - {p^2} = 0
That is p(2q−p)=0p\left( {2q - p} \right) = 0
On solving we get values of p as ‘p = 0p{\text{ }} = {\text{ }}0’ and ‘p = 2qp{\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 (x−h)2+(y−k)2=r2{\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}.