The equation of a line parallel to the line 3 | vedclass

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Question

(A) The equation of the circle is $x^{2} + y^{2} = 9$, which has center $(0, 0)$ and radius $r = 3$.Any line parallel to $3x + 4y = 0$ is of the form

Vedclass · Accepted Answer

$3x + 4y = 15$

Vedclass · Answer

(A) The equation of the circle is $x^{2} + y^{2} = 9$, which has center $(0, 0)$ and radius $r = 3$.Any line parallel to $3x + 4y = 0$ is of the form $3x + 4y = k$.The perpendicular distance from the center $(0, 0)$ to the line $3x + 4y - k = 0$ must be equal to the radius $r = 3$.Using the formula for distance from a point to a line: $\frac{|3(0) + 4(0) - k|}{\sqrt{3^{2} + 4^{2}}} = 3$.$\frac{|-k|}{\sqrt{9 + 16}} = 3$ $\Rightarrow \frac{|k|}{5} = 3$ $\Rightarrow |k| = 15$.So, $k = 15$ or $k = -15$.The lines are $3x + 4y = 15$ and $3x + 4y = -15$.For the line to touch the circle in the first quadrant, the intercept form $\frac{x}{5} + \frac{y}{3.75} = 1$ (for $k=15$) shows positive intercepts, which lies in the first quadrant.Thus, the required equation is $3x + 4y = 15$.

The equation of a line parallel to the line $3x + 4y = 0$ and touching the circle $x^{2} + y^{2} = 9$ in the first quadrant is:

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