Question

Find the polar of the point (4, - 1) with respect to the circle $2{x^2} + 2{y^2} = 11$ .

Answer 1

Hint:Polar of the points, is the locus of it and pole is a point with respect to which polar is determined. So, to find the polar of the circle, apply directly the formula of polar passing through the given point.

Complete step-by-step answer:
Complete step by step answer:
Given equation of the circle is:
$2{x^2} + 2{y^2} = 11$
We have to find the equation of its polar through the point ( 4 , -1 ). We know that, the equation of polar through point $({x_1},{y_1})$ will be obtained as,
$2x{x_1} + 2y{y_1} = 11$
Now, substitute the values of coordinates (4,-1) in the above equation, then we get,
$
  2x \times 4 + 2y \times ( - 1) = 11 \\
   \Rightarrow 8x - 2y = 11 \\
 $
Thus 8x – 2y = 11 is the required equation of line as polar of the given circle.

Additional Information:Polar through a point with respect to a circle is the straight line. If through a point P ( ${x_1},{y_1}$ ) , inside or outside the circle, then there be drawn any straight line to meet the given circle. It will be the locus of the point of intersection of the tangents at points. Thus polar is the locus of points.

Note:Equation to the polar of the point ( ${x_1},{y_1}$ ) with respect to the circle ${x^2} + {y^2} = {a^2}$ will be a straight line with given equation,
$x{x_1} + y{y_1} = {a^2}$
This formula can be directly applied to find the polar through the given pointy.
Also, the polar of the point (${x_1},{y_1}$ ) with respect to the circle is given by the equation below,
\[\left( {x{x_1}{\text{ }} + {\text{ }}y{y_1}{\text{ }} + g\left( {x + {x_1}} \right) + f\left( {y + {y_1}} \right){\text{ }} + c} \right) = 0\]
This equation is the same as the equation for the tangent to the circle at some given point ( ${x_1},{y_1}$ ) on the circle.