Question

P is a variable point on the line L = 0. Tangents are drawn to the circles ${x^2} + {y^2} = 4$ from P to touch it at Q and R. the parallelogram PQSR is completed. If P = (3, 4), then the coordinates of S are
$\left( a \right)\left( {\dfrac{{ - 46}}{{25}},\dfrac{{63}}{{25}}} \right)$
$\left( b \right)\left( {\dfrac{{ - 51}}{{25}},\dfrac{{ - 68}}{{25}}} \right)$
$\left( c \right)\left( {\dfrac{{ - 46}}{{25}},\dfrac{{68}}{{25}}} \right)$
$\left( d \right)\left( {\dfrac{{ - 68}}{{25}},\dfrac{{51}}{{25}}} \right)$

Answer 1

Hint: In this particular question first draw the pictorial representation of the above problem it will give us a clear picture of what we have to find out, then use the concept if all the sides of a parallelogram are equal then it a rhombus so use these concepts to reach the solution of the question.

Complete step-by-step solution:

The pictorial representation of the above problem is shown in the above figure.
From a point, P two tangents are drawn which touches the circle at point Q and R as shown in the above figure.
Now as we know that the tangents are drawn from an external point on the circle always have the same length.
So the length of PQ and PR has the same lengths........................ (1)
Now it is given that PQSR is a parallelogram i.e. opposite sides of PQSR are equal and parallel.
I.e. PR = QS and RS = PQ.............. (2)
Therefore, from (1) and (2) we have,
PQ = QS = SR = RP.
Now as we know that if all the lengths of the sides are equal and they are not perpendicular to each other then it is a Rhombus.
So, PQSR is a rhombus.
So the point S is a mirror image of the point P (3, 4) and passing from the center of the circle.
Let the coordinates of S is (a, b).
Now the equation of the circle is ${x^2} + {y^2} = 4$.
So the center and radius of the circle are (0, 0) and 2 units.
So the PS passing from the point (0, 0).
Now the equation of line passing from points $\left( {{x_1},{y_1}} \right){\text{ and }}\left( {{x_2},{y_2}} \right)$ is given as
$\left( {y - {y_1}} \right) = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}\left( {x - {x_1}} \right)$
Let, $\left( {{x_1},{y_1}} \right) = \left( {0,0} \right)$, and $\left( {{x_2},{y_2}} \right) = \left( {3,4} \right)$
So the equation of line is,
$ \Rightarrow 3y = 4x$
Now the equation is also passing from the point (a, b) so it satisfies the above equation so we have,
$ \Rightarrow 3b = 4a$
$ \Rightarrow \dfrac{a}{b} = \dfrac{3}{4}$
So the ratio of a to b is 3 : 4.
Now check all the options
Option (a)
(a, b) = $\left( {\dfrac{{ - 46}}{{25}},\dfrac{{63}}{{25}}} \right)$
So, $\dfrac{a}{b} = \dfrac{{\dfrac{{ - 46}}{{25}}}}{{\dfrac{{63}}{{25}}}} = \dfrac{{ - 46}}{{63}}$
So this is not the coordinates of S.
Option (b)
(a, b) = $\left( {\dfrac{{ - 51}}{{25}},\dfrac{{ - 68}}{{25}}} \right)$
So, $\dfrac{a}{b} = \dfrac{{\dfrac{{ - 51}}{{25}}}}{{\dfrac{{ - 68}}{{25}}}} = \dfrac{{51}}{{68}} = \dfrac{3}{4}$
So this is the coordinates of S.
Option (c)
(a, b) = $\left( {\dfrac{{ - 46}}{{25}},\dfrac{{68}}{{25}}} \right)$
So, $\dfrac{a}{b} = \dfrac{{\dfrac{{ - 46}}{{25}}}}{{\dfrac{{68}}{{25}}}} = \dfrac{{ - 46}}{{68}} = \dfrac{{ - 23}}{{34}}$
So this is not the coordinates of S.
Option (d)
(a, b) = $\left( {\dfrac{{ - 68}}{{25}},\dfrac{{51}}{{25}}} \right)$
So, $\dfrac{a}{b} = \dfrac{{\dfrac{{ - 68}}{{25}}}}{{\dfrac{{51}}{{25}}}} = \dfrac{{ - 68}}{{51}} = \dfrac{{ - 4}}{3}$
So this is not the coordinates of S.
Hence option (b) is the correct answer.

Note: Whenever we face such types of questions the key concept we have to remember is that the tangents drawn from an external point on the circle is always have same length, and always recall that the equation of line passing from points $\left( {{x_1},{y_1}} \right){\text{ and }}\left( {{x_2},{y_2}} \right)$ is given as, $\left( {y - {y_1}} \right) = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}\left( {x - {x_1}} \right)$