Question

The pole of the straight line 9x + y – 28 = 0 with respect to the circle is

- (3, 1)
- (1, 3)
- (3, – 1)
- (–3, 1)

Hint:

### Let P(h, k) be the pole of the circle x^{2} + y^{2} + 2gx + 2fy = 0 , then the equation of the polar line from the point is hx + ky + g(x + h) + f(y + k) + c= 0. We can find the equation of the polar by substituting the coordinates of the pole in the above equation.

## The correct answer is: (3, – 1)

### Let the coordinates of the pole be (h, k).

The equation of the circle is 2x^{2} + 2y^{2} - 3x + 5y - 7 = 0.

Dividing both sides of the equation by 2,we get

x^{2} + y^{2} - 3/2 x + 5/2 y - 7/2 = 0.

The equation of polar for the point having coordinates (h, k) is

h x + k y - 3/4 (x + h) + 5/4 (y+k) - 7/2 = 0 .

Simplifying the equation, it can be written as (4h - 3)x + (4k + 5)y + (-3h + 5k - 14).

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 coefficient of the line, we get

(4h - 3) /9 = (4k + 5) /1 = (-3h + 5k - 14) /-28

Taking the first and third expression, we get

(4h - 3) /9 = (-3h + 5k - 14) /-28 .

By cross multiplication, we get

Or, -28 (4h - 3) = 9 (-3h + 5k - 14)

Or, -112h + 84 = -27h + 45k - 126

Or, -112h + 27h - 45k = -126 - 84

Or, -85h - 45k = -210

Or, -5 ( 17h + 9k ) = -210

17h + 9k = 42 . . . (1)

Taking the second and third expression, we get

(4k + 5) /1 = ( -3h + 5k - 14) /-28

By cross multiplication, we get

-28 ( 4k + 5 ) = ( -3h + 5k - 14 )

-112k - 140 = -3h + 5k - 14

3h - 112k - 5k = -14 + 140

3h - 117k = 126 . . . (2)

By solving equation (1) × 13 and equation (2) .

h = 3

By putting the value of h in equation (1) , we get

K = -1 .

Therefore, the coordinates are (3, -1) .

Hence, the correct option is (a).

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.

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