The pole of the straight line 9x + y - 28 = 0 | vedclass

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(C) The equation of the circle is $2x^2 + 2y^2 - 3x + 5y - 7 = 0$. Dividing by $2$,we get $x^2 + y^2 - \frac{3}{2}x + \frac{5}{2}y - \frac{7}{2} = 0$.

Vedclass · Accepted Answer

$(3, -1)$

Vedclass · Answer

(C) The equation of the circle is $2x^2 + 2y^2 - 3x + 5y - 7 = 0$. Dividing by $2$,we get $x^2 + y^2 - \frac{3}{2}x + \frac{5}{2}y - \frac{7}{2} = 0$.The general form of the polar of a point $(x_1, y_1)$ with respect to the circle $x^2 + y^2 + 2gx + 2fy + c = 0$ is $xx_1 + yy_1 + g(x + x_1) + f(y + y_1) + c = 0$.Comparing with our circle,$g = -\frac{3}{4}$,$f = \frac{5}{4}$,and $c = -\frac{7}{2}$.The equation of the polar is $xx_1 + yy_1 - \frac{3}{4}(x + x_1) + \frac{5}{4}(y + y_1) - \frac{7}{2} = 0$.Rearranging terms: $x(x_1 - \frac{3}{4}) + y(y_1 + \frac{5}{4}) - \frac{3}{4}x_1 + \frac{5}{4}y_1 - \frac{7}{2} = 0$.This must be identical to the given line $9x + y - 28 = 0$,or $x + \frac{1}{9}y - \frac{28}{9} = 0$.Comparing coefficients: $\frac{x_1 - 3/4}{1} = \frac{y_1 + 5/4}{1/9} = \frac{-3/4x_1 + 5/4y_1 - 7/2}{-28/9} = k$.From $x_1 - 3/4 = k$ and $y_1 + 5/4 = k/9$,we get $x_1 = k + 3/4$ and $y_1 = k/9 - 5/4$.Substituting into the constant term ratio: $\frac{-3/4(k + 3/4) + 5/4(k/9 - 5/4) - 7/2}{-28/9} = k$.Solving for $k$,we find $k = 9/4$. Thus,$x_1 = 9/4 + 3/4 = 3$ and $y_1 = (9/4)/9 - 5/4 = 1/4 - 5/4 = -1$.The pole is $(3, -1)$.

The pole of the straight line $9x + y - 28 = 0$ with respect to the circle $2x^2 + 2y^2 - 3x + 5y - 7 = 0$ is:

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