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Question: Find the condition that straight line \( \dfrac{k}{r} = A\cos \theta + B\sin \theta \) may touch the...

Find the condition that straight line kr=Acos⁡θ+Bsin⁡θ\dfrac{k}{r} = A\cos \theta + B\sin \theta may touch the circle r=2acos⁡θr = 2a\cos \theta .

Explanation

Solution

Hint : First, simplify the equation of the line to its general form, and then simplify the equation of the circle to the general form too. Find the relation between the two equations. Use the distance formula for calculating the distance between the centre of the circle and the given line to find the required condition.

Complete step-by-step answer :
We are given that the equation of the line is,
kr=Acos⁡θ+Bsin⁡θ ⇒k=Arcos⁡θ+Brsin⁡θ  \dfrac{k}{r} = A\cos \theta + B\sin \theta \\\ \Rightarrow k = Ar\cos \theta + Br\sin \theta \\\
Let, rcos⁡θ=xr\cos \theta = x and rsin⁡θ=yr\sin \theta = y
We get, k=Ax+Byk = Ax + By
Equation of circle is r=2acos⁡θr = 2a\cos \theta
Multiplying rr on both sides, we get –
r2=2arcos⁡θ{r^2} = 2ar\cos \theta
We know that
x=rcos⁡θx = r\cos \theta
Squaring both sides
x2=r2cos⁡2θ{x^2} = {r^2}{\cos ^2}\theta
Also, y=rsin⁡θy = r\sin \theta
Squaring both sides
y2=r2sin⁡2θ{y^2} = {r^2}{\sin ^2}\theta
Adding the above two equations, we get –
⇒x2+y2=r2(cos⁡2θ+sin⁡2θ) ⇒x2+y2=r2   \Rightarrow {x^2} + {y^2} = {r^2}({\cos ^2}\theta + {\sin ^2}\theta ) \\\ \Rightarrow {x^2} + {y^2} = {r^2} \;
Putting the value obtained in the simplified equation of the circle, we get –
⇒x2+y2=2ax ⇒x2−2ax+y2=0   \Rightarrow {x^2} + {y^2} = 2ax \\\ \Rightarrow {x^2} - 2ax + {y^2} = 0 \;
Adding a2{a^2} on both sides, we get –
⇒x2−2ax+a2+y2=a2 ⇒(x)2+(a)2−2×a×x+y2=a2 ⇒(x−a)2+y2=a2   \Rightarrow {x^2} - 2ax + {a^2} + {y^2} = {a^2} \\\ \Rightarrow {(x)^2} + {(a)^2} - 2 \times a \times x + {y^2} = {a^2} \\\ \Rightarrow {(x - a)^2} + {y^2} = {a^2} \;
We know that the general equation of a circle is (x−h)2+(y−k)2=r2{(x - h)^2} + {(y - k)^2} = {r^2}
Where (h,k)(h,k) is the coordinates of the centre of the circle and rr is the radius of the circle.
On comparing this equation with the simplified equation of the circle, we see that the coordinates of the centre of the given circle are (a,0)(a,0) and its radius is aa units.
When a line touches a circle at any point, the line becomes the tangent to the circle at that point.
As the distance of any point on the circle from its centre is equal to its radius, so the distance of the line from the centre of the circle is aa .
Also, we know that the perpendicular distance of a line having equation ax+by+c=0ax + by + c = 0 from a point (x0,y0)({x_0},{y_0}) is given by the formula –
⇒d=∣ax0+by0+c∣a2+b2\Rightarrow d = \dfrac{{\left| {a{x_0} + b{y_0} + c} \right|}}{{\sqrt {{a^2} + {b^2}} }}
⇒∣A(a)+b(0)−k∣A2+B2=a ⇒∣Aa−k∣=aA2+B2   \Rightarrow \dfrac{{\left| {A(a) + b(0) - k} \right|}}{{\sqrt {{A^2} + {B^2}} }} = a \\\ \Rightarrow \left| {Aa - k} \right| = a\sqrt {{A^2} + {B^2}} \;
To remove the square root and modulus, square both sides of the above equation,
⇒(Aa−k)2=a2(A2+B2) ⇒A2a2+k2−2Aak=a2A2+a2B2 ⇒k2−2aAk−a2B2=0   \Rightarrow {(Aa - k)^2} = {a^2}({A^2} + {B^2}) \\\ \Rightarrow {A^2}{a^2} + {k^2} - 2Aak = {a^2}{A^2} + {a^2}{B^2} \\\ \Rightarrow {k^2} - 2aAk - {a^2}{B^2} = 0 \;
Thus the condition that the given line touches the given circle is,
k2−2aAk−a2B2=0{k^2} - 2aAk - {a^2}{B^2} = 0
So, the correct answer is “ k2−2aAk−a2B2=0{k^2} - 2aAk - {a^2}{B^2} = 0 ”.

Note : As the tangent at a point is perpendicular to the line joining the centre and the point, the distance of the given line from the centre of the given circle is calculated by the formula of finding the perpendicular distance of a point from a line.