
The pole of the straight line is \[9x+y-28=0\] with respect to the circle \[2{{x}^{2}}+2{{y}^{2}}-3x+5y-7=0\], is
\[\begin{align}
& \left( A \right)\left( 3,1 \right) \\
& \left( B \right)\left( 1,3 \right) \\
& \left( C \right)\left( 3,-1 \right) \\
& \left( D \right)\left( -3,1 \right) \\
\end{align}\]
Answer
540k+ views
Hint: Let \[P(h,k)\] be the pole of the circle \[{{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0\], then the equation of the polar line from the point is \[hx+ky+g\left( x+h \right)+f\left( y+k \right)+c=0\]. We can find the equation of the polar by substituting the coordinates of the pole in the above equation. It should be noted that the coefficient of the square terms in the equation of the circle is one, so if it is not one, make it by dividing by the coefficient.
Complete step by step answer:
Let the coordinates of the pole be \[\left( h,k \right)\]. The equation of the circle is \[2{{x}^{2}}+2{{y}^{2}}-3x+5y-7=0\], dividing both sides of the equation by 2, we get \[{{x}^{2}}+{{y}^{2}}-\dfrac{3}{2}x+\dfrac{5}{2}y-\dfrac{7}{2}=0\].
The equation of polar for the point having coordinates \[\left( h,k \right)\] is \[hx+ky-\dfrac{3}{4}\left( x+h \right)+\dfrac{5}{4}\left( y+k \right)-\dfrac{7}{2}=0\]. Simplifying the equation, it can be written as \[\left( 4h-3 \right)x+\left( 4k+5 \right)y+\left( -3h+5k-14 \right)\].
We are given that the equation of the polar is \[9x+y-28=0\]. So the above equation and this equation represent the same line. Comparing the coefficients of the line, we get
\[\dfrac{\left( 4h-3 \right)}{9}=\dfrac{\left( 4k+5 \right)}{1}=\dfrac{\left( -3h+5k-14 \right)}{-28}\]
Taking the first and third expressions, we get \[\dfrac{\left( 4h-3 \right)}{9}=\dfrac{\left( -3h+5k-14 \right)}{-28}\]. Taking the second and third and expression, we get \[\dfrac{\left( 4k+5 \right)}{1}=\dfrac{\left( -3h+5k-14 \right)}{-28}\].
By solving the above two equations, we get \[h=3\And k=-1\]. Hence the coordinates of the pole are \[\left( 3,-1 \right)\].
We can also show the pole and polar on the graph as follows,
Complete step by step answer:
Let the coordinates of the pole be \[\left( h,k \right)\]. The equation of the circle is \[2{{x}^{2}}+2{{y}^{2}}-3x+5y-7=0\], dividing both sides of the equation by 2, we get \[{{x}^{2}}+{{y}^{2}}-\dfrac{3}{2}x+\dfrac{5}{2}y-\dfrac{7}{2}=0\].
The equation of polar for the point having coordinates \[\left( h,k \right)\] is \[hx+ky-\dfrac{3}{4}\left( x+h \right)+\dfrac{5}{4}\left( y+k \right)-\dfrac{7}{2}=0\]. Simplifying the equation, it can be written as \[\left( 4h-3 \right)x+\left( 4k+5 \right)y+\left( -3h+5k-14 \right)\].
We are given that the equation of the polar is \[9x+y-28=0\]. So the above equation and this equation represent the same line. Comparing the coefficients of the line, we get
\[\dfrac{\left( 4h-3 \right)}{9}=\dfrac{\left( 4k+5 \right)}{1}=\dfrac{\left( -3h+5k-14 \right)}{-28}\]
Taking the first and third expressions, we get \[\dfrac{\left( 4h-3 \right)}{9}=\dfrac{\left( -3h+5k-14 \right)}{-28}\]. Taking the second and third and expression, we get \[\dfrac{\left( 4k+5 \right)}{1}=\dfrac{\left( -3h+5k-14 \right)}{-28}\].
By solving the above two equations, we get \[h=3\And k=-1\]. Hence the coordinates of the pole are \[\left( 3,-1 \right)\].
We can also show the pole and polar on the graph as follows,
Hence, the correct option is C.
Note: Pole and polar are one of the important parts of the circle, many questions can be based on these parts. It should be remembered that while writing the equation of polar for a pole, the coefficients of square terms in the equation of the circle is one. If the equation of the circle does not have coefficients as one, then make it one before solving the question.
Note: Pole and polar are one of the important parts of the circle, many questions can be based on these parts. It should be remembered that while writing the equation of polar for a pole, the coefficients of square terms in the equation of the circle is one. If the equation of the circle does not have coefficients as one, then make it one before solving the question.
Recently Updated Pages
Why are manures considered better than fertilizers class 11 biology CBSE

Find the coordinates of the midpoint of the line segment class 11 maths CBSE

Distinguish between static friction limiting friction class 11 physics CBSE

The Chairman of the constituent Assembly was A Jawaharlal class 11 social science CBSE

The first National Commission on Labour NCL submitted class 11 social science CBSE

Number of all subshell of n + l 7 is A 4 B 5 C 6 D class 11 chemistry CBSE

Trending doubts
10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

