The equation of a line passing through (7, 4) | vedclass

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(D) The equation of the circle is $x^2 + y^2 - 6x + 4y - 3 = 0$. The center is $(3, -2)$ and the radius $r = \sqrt{3^2 + (-2)^2 - (-3)} = \sqrt{9 + 4

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$(A)$ and $(C)$ both

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(D) The equation of the circle is $x^2 + y^2 - 6x + 4y - 3 = 0$. The center is $(3, -2)$ and the radius $r = \sqrt{3^2 + (-2)^2 - (-3)} = \sqrt{9 + 4 + 3} = 4$.Let the line passing through $(7, 4)$ be $y - 4 = m(x - 7)$,which simplifies to $mx - y + (4 - 7m) = 0$.The perpendicular distance from the center $(3, -2)$ to the line must equal the radius $r = 4$.$\frac{|m(3) - (-2) + 4 - 7m|}{\sqrt{m^2 + (-1)^2}} = 4$$\frac{|6 - 4m|}{\sqrt{m^2 + 1}} = 4$Squaring both sides: $(6 - 4m)^2 = 16(m^2 + 1)$$36 - 48m + 16m^2 = 16m^2 + 16$$-48m = -20 \Rightarrow m = \frac{20}{48} = \frac{5}{12}$.Substituting $m = 5/12$ into the line equation: $y - 4 = \frac{5}{12}(x - 7)$ $\Rightarrow 12y - 48 = 5x - 35$ $\Rightarrow 5x - 12y + 13 = 0$.Also,the vertical line $x = 7$ passes through $(7, 4)$. The distance from $(3, -2)$ to $x - 7 = 0$ is $|3 - 7| = 4$,which equals the radius. Thus,$x - 7 = 0$ is also a tangent.Therefore,both $(A)$ and $(C)$ are correct.

The equation of a line passing through $(7, 4)$ and touching the circle $x^2 + y^2 - 6x + 4y - 3 = 0$ is:

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