From a point P on the line 4x - 3y = 6,two ta | vedclass

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(A) The given circle is $x^2 + y^2 - 6x - 4y + 4 = 0$. Its center is $C(3, 2)$ and radius $r = \sqrt{3^2 + 2^2 - 4} = \sqrt{9 + 4 - 4} = 3$. Let the a

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$(6, 6)$

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(A) The given circle is $x^2 + y^2 - 6x - 4y + 4 = 0$. Its center is $C(3, 2)$ and radius $r = \sqrt{3^2 + 2^2 - 4} = \sqrt{9 + 4 - 4} = 3$. Let the angle between the tangents be $2\alpha$. Then $2\alpha = 	an^{-1}\left(\frac{24}{7}ight)$,so $	an(2\alpha) = \frac{24}{7}$. Using $	an(2\alpha) = \frac{2	an\alpha}{1 - 	an^2\alpha} = \frac{24}{7}$,we get $12	an^2\alpha + 7	an\alpha - 12 = 0$. Solving this,we find $	an\alpha = 3/4$. In the right-angled triangle $	riangle PAC$,$\sin\alpha = \frac{r}{CP} = \frac{3}{CP}$. Since $	an\alpha = 3/4$,we have $\sin\alpha = 3/5$. Thus,$\frac{3}{CP} = \frac{3}{5} \Rightarrow CP = 5$. If $P = (h, k)$,then $(h-3)^2 + (k-2)^2 = 5^2 = 25$,which simplifies to $h^2 + k^2 - 6h - 4k - 12 = 0$. Since $P$ lies on $4x - 3y = 6$,$h = \frac{6 + 3k}{4}$. Substituting this into the circle equation gives $k^2 - 4k - 12 = 0$,so $(k-6)(k+2) = 0$. If $k = 6$,$h = 6$. If $k = -2$,$h = 0$. Thus,$P$ can be $(6, 6)$ or $(0, -2)$.

From a point $P$ on the line $4x - 3y = 6$,two tangents are drawn to the circle $x^2 + y^2 - 6x - 4y + 4 = 0$. If the angle between these tangents is $\tan^{-1}\left(\frac{24}{7}\right)$,then $P$ can be:

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