MATH -127 (Coordinate Geometry)
Chapter -3The Circle
• The pole is outside the circle.
• The pole is inside the circle
• The pole is on the polar
♦ Polar and Pole: If through a point P, we draw a straight line to meet a circle or conic at two points Q and R, the locus of the points of intersection of the tangents drawn at Q, R is a line known as the polar and the point P is the pole of the polar.
locus is the line that is made of a 'set of points'.
locus is the line that is made of a 'set of points'.
Question 1: Find the pole of the line
\(2x - y = 6\)
with respect to the circle
\(5x^2 + 5 y^2 = 9\)
Solution: given that,
\(2x - y = 6 \tag{1.1}\label{eq:11.}\)
\(5x^2 + 5 y^2 = 9 \tag{1.2}\label{eq:1.2}\)
The equation of the polar with respect to the circle,
\(5x^2 + 5 y^2 = 9 ~\textrm{of P} (x_1, y_1) ~\textrm{is }~ 5xx_1 + 5yy_1 = 9 \tag{1.3}\label{eq:1.3}\)
Since equation (1.3) is identical with the given line, we have,
\begin{align*}
& \frac{5 x_1}{2} = \frac{5y_1}{- 2} = \frac{9}{6} \\
\implies & x_1 = \frac{3}{5},~~~~~ y_1 = - \frac{3}{10}
\end{align*}
The required pole is
\( \left(\frac{3}{5}, \frac{- 3}{10}\right)\)
Question 2: Show that the equation to the pair of tangents drawn from the origin to the circle
\(x^2 + y^2 + 2gx + 2fy + c = 0 ~\textrm{is }~ c( x^2 + y^2) = ( gx + fy )^2\)
Solution The equation to the pair of tangents is,
\begin{align*}
& SS_1 = T^2 \\
\implies & ( x^2 + y^2 + 2gx + 2fy + c ) c = [x \times 0 + y \times 0 + g(x + 0) + f(y + 0) + c]^2\\
\implies & c ( x^2 + y^2 ) + c ( 2gx + 2fy + c) = (gx + fy + c)^2 + ( gx + fy )^2 + 2 (gx + fy ) c + c^2\\
\implies & c ( c^2 + y^2) = (gx + fy)^2
\end{align*}
Question 3: Prove that if the polar of a point P with respect to the circle
\( x^2 + y^2 = 37\)
touches the circle
\((x - 3)^2 + (y + 2 )^2 = 25\)
then the locus of p must be a cone.
Solution: Given circles are
\( x^2 + y^2 = 37 \tag{3.1}\label{eq:3.1}\) \((x - 3)^2 + (y + 2 )^2 = 25 \tag{3.2}\label{eq:3.2}\)
Let, \(P(x_1, y_1)\) be the pole.
Then the polar of P with respect to the circle
\(x^2 + y^2 = 37\)
is
\(xx_1 + yy_1 = 37 \tag{3.3}\label{eq:3.3}\)
Since (3.1) touches (3.2) we have,
$$ \frac{3x_1 - 2y_1 - 37}{\sqrt{{x_1}^2 + {y_1}^2}} $$
Since length of the perpendicular from \((3, - 2)\) to (3.1) is equal to the radius of the circle. so that,
\begin{align*}
& (3x_1 - 2y_1 - 37)^2 = 25 ({x_1}^2 + {y_1}^2) \\
\implies & 16 {x_1}^2 + 21 {y_1}^2 + 12xy + 222x_1 - 148 y_1 - 1369 = 0
\end{align*}
\(\therefore\) the required locus is,
\(16 {x_1}^2 + 21 {y_1}^2 + 12xy + 222x_1 - 148 y_1 - 1369 = 0\)
Question 4: Find the pole of the straight line
\( 9x + y = 28\)
with respect to the circle
\(2 x^2 + 2 y^2 - 3 x + 5 y - 7 = 0 \)
Solution: The given equations are
\( 9x + y = 28 \tag{4.1}\label{eq:4.1}\) \(2 x^2 + 2 y^2 - 3 x + 5 y - 7 = 0 \tag{4.2}\label{eq:4.2}\)
If \((x_1, y_1)\) be the required point the line (4.1) must coincide with the polar of \((x_1, y_1)\) whose equation is,
\begin{align*}
& 2xx_1 + 2yy_1 - \frac{3}{2}(x_1 + x ) + \frac{5}{2}(y_1 + y) - 7 = 0 \\
\implies & x ( 4 x_1 - 3 ) + y (4y_1 + 5 ) - 3 x_1 + 4 y_1 - 14 = 0 \tag{4.3}\label{eq:4.3}
\end{align*}
Since (4.1) and (4.3) are the same, we have
$$ \frac{4x_1 - 3}{9} = \frac{4y_1 + 5}{1} = \frac{- 3x_1 + 5 y_1 - 14}{- 28} $$
Hence,
\begin{align*}
& x_1 = 9 y_1 + 14 \tag{4.4}\label{eq:4.4} \\
& 3x_1 - 117 y_1 = 126 \tag{4.5}\label{eq:4.5}
\end{align*}
solving equation (4.4) and (4.5) we get, \(x_1 = 3, ~~ y_1 = - 1\)
\(\therefore\) The required point is \((3,- 1)\)
Question 5: Find the polar of the point \((4,- 1)\) with respect to the circle
\(2 x^2 + 2 y^2 = 11\)
Solution: The equation of the polar of the circle is,
\begin{align*}
& 2xx_1 + 2 yy_1 = 11\\
\implies & 2\times 4x + 2\times (- 1) y = 11 \\
\implies & 8x - 2 y = 11
\end{align*}
Question 6: Find the polar of the point \((- 2, 3)\) with respect to the circle
\( x^2 + y^2 - 4x - 6y + 5 = 0\)
Solution: The equation of the polar of the circle is,
\begin{align*}
& xx_1 + yy_1 - \frac{4}{2} ( x + x_1) - \frac{6}{2} (y + y_1) + 5 = 0 \\
\implies & - 2x + 3 y - 2 (x - 2) - 3 (y + 3) + 5 = 0 \\
\implies & - 2x + 3y - 2x + 4 - 3y - 9 + 5 = 0 \\
\implies & - 4x = 0\\
\therefore & ~ x = 0
\end{align*}
Question 7: Find the pole of the straight line
\(2x + y + 12 = 0\)
with respect to the circle
\(x^2 + y^2 - 4x + 3y - 1 = 0\)
Solution: The given equations are,
\(2x + y + 12 = 0 \tag{7.1}\label{eq:7.1}\) \(x^2 + y^2 - 4x + 3y - 1 = 0 \tag{7.2}\label{eq:7.2}\)
Let \((x_1, y_1)\) be the pole so the line (7.1) must coincide with the point \((x_1, y_1)\) whose equation is,
\begin{align*}
& xx_1 + yy_1 - \frac{4}{2}(x + x_1) + \frac{3}{2}(y + y_1) - 1 = 0\\
\implies & x ( x_1 - 2) + y ( y_1 + \frac{3}{2}) - 2 x_1 + \frac{3}{2} y_1 - 1 = 0 \tag{7.3}\label{eq:7.3}
\end{align*}
Since (7.1) and (7.3) are same we have,
$$ \frac{x_1 - 2}{2} = \frac{ y_1 + \frac{3}{2}}{1} = \frac{2x_1 - \frac{3}{2}y_1 + 1}{- 12} $$
Taking 1st and 3rd part we get,
\begin{align*}
& - 6x_1 + 12 = 2 x_1 - \frac{3}{2}y_1 + 1 \\
\implies & - 8x_1 + 12 + \frac{3}{2} y_1 + 1 = 0 \\
\implies & - 16 x_1 + 3 y_1 + 26 = 0 \tag{7.4}\label{eq:7.4}
\end{align*}
Taking 2nd and 3rd part we get,
\begin{align*}
& y_1 + \frac{3}{2} = \frac{2x_1 - \frac{3}{2}y_1 + 1}{- 12}\\
\implies & - 12 y_1 - 18 = 2 x_1 - \frac{3}{2} y_1 + 1 \\
\implies & - 24y_1 - 36 = 4x_1 - 3 y_1 + 2\\
\implies & - 24 y_1 - 3 y_1 - 36 - 2 - 4 x_1 = 0 \\
\implies & 4 x_1 + 27 y_1 + 38 = 0 \tag{7.5}\label{eq:7.5}
\end{align*}
Solving (7.4) and (7.5) we get,
\begin{align*}
& x_1 = \frac{49}{37}\\
& y_1 = - \frac{178}{111}
\end{align*}
\(\therefore\) The required pole is
\((x_1, y_1) \equiv \left(\frac{49}{37}, - \frac{ 178}{111}\right) \)
Question 8: If the polar of the point P with respect to the circle
\(x^2 + y^2 - 2bx = 0\)
touches the circle
\(x^2 + y^2 = a^2\)
. Find the locus of P.
Solution: The given equations are
\(x^2 + y^2 - 2bx = 0 \tag{8.1}\label{eq:8.1}\) \(x^2 + y^2 = a^2 \tag{8.2}\label{eq:8.2}\)
Let P\((x_1, y_1)\) be the pole, then the polar of the circle
\(x^2 + y^2 - 2bx = 0 \)
is
\(xx_1 + yy_1 - 2b(x + x_1) = 0 \tag{8.3}\label{eq:8.3}\)
Since (8.1) touches (8.2) we have,
\begin{align*}
& \frac{x_1 \times 0 + y_1 \times 0 - 2b(0 + x_1)}{\sqrt{{x_1}^2 + {y_1}^2}} = 1 \textrm{ [since, length of the perpendicular is equal to the radius]} \\
& - 2bx_1 = a \sqrt{ {x_1}^2 + {y_1}^2} \\
& 4 b^2 {x_1}^2 = a^2 ( {x_1}^2 + {y_1}^2) \\
& a^2 {x_1}^2 + a^2 {y_1}^2 - 4b^2{x_1}^2 = 0 \\
( a^2 - 4 b^2) {x_1}^2 + a^2 {y_1}^2 = 0
\end{align*}
Find the equation of the direct common tangents to the circles
\(x^2 + y^2 = 16 \textrm{ and } x^2 + y^2 + 6 x - 78 y = 0\)
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