The pole of the line x-5y-7=0 with respect to | vedclass

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(C) The equation of the circle is $S \equiv x^2+y^2-2x+4y+1=0$. The centre $C$ is $(1, -2)$.For a line $lx+my+n=0$ and circle $x^2+y^2+2gx+2fy+c=0$,th

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$\sqrt{a^3+b^3-1}$

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(C) The equation of the circle is $S \equiv x^2+y^2-2x+4y+1=0$. The centre $C$ is $(1, -2)$.For a line $lx+my+n=0$ and circle $x^2+y^2+2gx+2fy+c=0$,the pole $(x_1, y_1)$ satisfies $\frac{x_1+g}{l} = \frac{y_1+f}{m} = \frac{gx_1+fy_1+c}{-n}$.Here $g=-1, f=2, c=1, l=1, m=-5, n=-7$.So,$\frac{a-1}{1} = \frac{b+2}{-5} = \frac{-a+2b+1}{7}$.Solving $\frac{a-1}{1} = \frac{b+2}{-5}$,we get $5a-5 = -b-2 \Rightarrow 5a+b=3$.Solving $\frac{a-1}{1} = \frac{-a+2b+1}{7}$,we get $7a-7 = -a+2b+1$ $\Rightarrow 8a-2b=8$ $\Rightarrow 4a-b=4$.Adding the two equations: $9a=7 \Rightarrow a=7/9$ and $b=4a-4 = 28/9 - 36/9 = -8/9$.Wait,re-evaluating the pole: $a-1 = k, b+2 = -5k, -a+2b+1 = -7k$.$-k-1 + 2(-5k-2) + 1 = -7k$ $\Rightarrow -k-1-10k-4+1 = -7k$ $\Rightarrow -11k-4 = -7k$ $\Rightarrow 4k = -4$ $\Rightarrow k=-1$.Thus,$a-1 = -1 \Rightarrow a=0$ and $b+2 = 5 \Rightarrow b=3$.The pole $P$ is $(0, 3)$.The distance $PC = \sqrt{(0-1)^2 + (3-(-2))^2} = \sqrt{1^2 + 5^2} = \sqrt{26}$.Checking option $C$: $\sqrt{a^3+b^3-1} = \sqrt{0^3+3^3-1} = \sqrt{27-1} = \sqrt{26}$.Therefore,$PC = \sqrt{a^3+b^3-1}$.

The pole of the line $x-5y-7=0$ with respect to the circle $S \equiv x^2+y^2-2x+4y+1=0$ is $P(a, b)$. If $C$ is the centre of the circle $S=0$,then $PC=$

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