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Question: Find the locus of point of intersection of the lines \[x\cos \alpha + y\sin \alpha = a\] and \[x\...

Find the locus of point of intersection of the lines
xcos⁡α+ysin⁡α=ax\cos \alpha + y\sin \alpha = a and xsin⁡α−ycos⁡α=bx\sin \alpha - y\cos \alpha = b, where α\alpha is a parameter.
A.x2−y2=a2+b2{x^2} - {y^2} = {a^2} + {b^2}
B.x2−y2=a2−b2{x^2} - {y^2} = {a^2} - {b^2}
C.x2+y2=a2−b2{x^2} + {y^2} = {a^2} - {b^2}
D.x2+y2=a2+b2{x^2} + {y^2} = {a^2} + {b^2}

Explanation

Solution

Here, we are required to find the locus of points of intersection of the given lines. We will assume the coordinate of the point of intersection of the given lines to be some variable. Then we will substitute the values of the coordinate in the given equation of line. We will then square and add these equations and simplify it to find the required locus of the point of intersection of the two given equations.

Complete step-by-step answer:
Let the point of intersection of the lines xcos⁡α+ysin⁡α=ax\cos \alpha + y\sin \alpha = a and xsin⁡α−ycos⁡α=bx\sin \alpha - y\cos \alpha = b be P(h,k)P\left( {h,k} \right).
Since, this point is a point of intersection, hence, this will satisfy both the lines, i.e. at point P(h,k)P\left( {h,k} \right), the given lines will be:
hcos⁡α+ksin⁡α=ah\cos \alpha + k\sin \alpha = a……………………….(1)\left( 1 \right)
And
hsin⁡α−kcos⁡α=bh\sin \alpha - k\cos \alpha = b……………………….(2)\left( 2 \right)
Now, squaring and adding both the equation (1)\left( 1 \right) and (2)\left( 2 \right), we get
(hcos⁡α+ksin⁡α)2+(hsin⁡α−kcos⁡α)2=a2+b2{\left( {h\cos \alpha + k\sin \alpha } \right)^2} + {\left( {h\sin \alpha - k\cos \alpha } \right)^2} = {a^2} + {b^2}
Now, computing the square on the expression using the formula (a±b)2=a2±2ab+b2{\left( {a \pm b} \right)^2} = {a^2} \pm 2ab + {b^2}, we get
⇒h2cos⁡2α+2hcos⁡α.ksin⁡α+k2sin⁡2α+h2sin⁡2α−2hsin⁡α.kcos⁡α+k2cos⁡2α=a2+b2\Rightarrow {h^2}{\cos ^2}\alpha + 2h\cos \alpha .k\sin \alpha + {k^2}{\sin ^2}\alpha + {h^2}{\sin ^2}\alpha - 2h\sin \alpha .k\cos \alpha + {k^2}{\cos ^2}\alpha = {a^2} + {b^2}
Factoring out common terms, we get
⇒h2(sin⁡2α+cos⁡2α)+k2(sin⁡2α+cos⁡2α)=a2+b2\Rightarrow {h^2}\left( {{{\sin }^2}\alpha + {{\cos }^2}\alpha } \right) + {k^2}\left( {{{\sin }^2}\alpha + {{\cos }^2}\alpha } \right) = {a^2} + {b^2}
Now, as we know (sin⁡2α+cos⁡2α)=1\left( {{{\sin }^2}\alpha + {{\cos }^2}\alpha } \right) = 1
Hence,
⇒h2(1)+k2(1)=a2+b2\Rightarrow {h^2}\left( 1 \right) + {k^2}\left( 1 \right) = {a^2} + {b^2}
⇒h2+k2=a2+b2\Rightarrow {h^2} + {k^2} = {a^2} + {b^2}
Since the point of intersection P(h,k)P\left( {h,k} \right) satisfies the given lines, so x2+y2=a2+b2{x^2} + {y^2} = {a^2} + {b^2} is the locus of the point of intersection of the given lines.
This forms a circle with radius a2+b2{a^2} + {b^2} and centre at the origin.
Therefore, option D is the correct answer and x2+y2=a2+b2{x^2} + {y^2} = {a^2} + {b^2} is the required answer.

Note: In mathematics, a locus is a set of all points that satisfy some properties which usually results as a curve or a surface. We also know that when two or more lines intersect each other at a point, that point is known as the point of intersection. In this question, the locus of the point of intersection of the given lines forms a circle with origin as its centre and with the radius a2+b2{a^2} + {b^2}.