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$x^2 + y^2 = 4$ and $3x + 4y = 12$

line from origin with gradient $\frac{4}{3}$ intersect circle at $(\frac{6}{5} , \frac{8}{5}) $. intersect $3x + 4y = 12$ at $(\frac{36}{25}, \frac{36}{25})$. so distance is $\sqrt{{(\frac{36}{25}- \frac{6}{5})}^2 + {(\frac{36}{25}- \frac{8}{5})}^2} = \frac{\sqrt{52}}{25}$

but my answer is wrong. Where am i wrong?

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    $\begingroup$ The line intyersects the circle at $(\frac65,\frac85)$ for me. $\endgroup$ – Bernard Jul 26 '19 at 17:15
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    $\begingroup$ you still have the point where the line from the origin (with slope $\frac43$) intersects $3x+4y=12$ wrong $\endgroup$ – J. W. Tanner Jul 26 '19 at 17:20
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No need to find the intersection, that is no need for quadratics. The normal to $y=-\frac34x+3$ through the origin is given by the equation $y=\frac43x$; determine the intersection point of both to find that the distance from the line to the origin is $\frac{3}{\sqrt{1+\left(-\frac34\right)^2}}=\frac{12}{5}$. Hence the distance from the circle to the line is $\frac{12}{5}-2$.

NB: In general the distance from $y=mx+b$ to the origin is $\frac{|b|}{\sqrt{1+m^2}}$

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  • $\begingroup$ yes. why i didn't notice it.. $\endgroup$ – Lifeforbetter Jul 26 '19 at 17:38
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    $\begingroup$ We are all only human beings ... $\endgroup$ – Michael Hoppe Jul 26 '19 at 17:38
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Your idea is correct.

You need to redo your calculation of intersects.

You have the point $(\frac {6}{5}, \frac {6}{5})$ on the line $y=\frac {4}{3} x$ which does not make sense.

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  • $\begingroup$ yes, it doesn't make sense. thank you.. $\endgroup$ – Lifeforbetter Jul 26 '19 at 17:26
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Another way:

Using coordinate geometry,

Any point on the circle $P(2\cos t,2\sin t)$

Distance of $P$ from the line $$\dfrac{|3(2\cos t)+4(2\sin t)-12|}5$$

Now $-\sqrt{6^2+8^2}\le 6\cos t+8\sin t\le?$

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  • $\begingroup$ So answer must be $\frac{2}{5}$ I guess? $\endgroup$ – user572457 Jul 26 '19 at 17:24
  • $\begingroup$ ok, thank you.. but not really got it.. $\endgroup$ – Lifeforbetter Jul 26 '19 at 17:43
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    $\begingroup$ @Lifeforbetter for any line $ax+by+c=0$ and any point (h,k), the perpendicular of the line from (h,k) is $\frac{|ah+bk+c|}{\sqrt{a^2+b^2}}$ $\endgroup$ – user572457 Jul 26 '19 at 17:56
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    $\begingroup$ @Lifeforbetter any point on the circle is of the form $(r\cos(t),r\sin(t))$. So we have to minimize the perpendicular distance. That is what the answer does. $\endgroup$ – user572457 Jul 26 '19 at 17:58
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    $\begingroup$ @LifeForBetter $$-10-12\le 6\cos t+8\sin t-12\le10-12$$ right? $\endgroup$ – lab bhattacharjee Jul 26 '19 at 18:21
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Let $$x_c^2+y_c^2=4$$ and $$3x_s+4y_s=12$$ then $$d=\sqrt{(x_c-x_s)^2+(y_c-y_s)^2}$$ you can eliminate two variables.

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  • $\begingroup$ yes. thank you.. $\endgroup$ – Lifeforbetter Jul 26 '19 at 17:26
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Hint

$3x=12-4y=4(3-y)$

WLOG any point on the line $P(4m,3-3m)$

$\dfrac43=\dfrac{3-3m-0}{4m-0}$

$m=?$

Can you find the distance of $P$ from the origin

Subtract the length of the radius from the distance

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Let $c$ be a real number such that $3x+4y=c$ has common points $(x,y)$ with the circle and parallel to $3x+4y=12$.

Thus, by C-S we obtain: $$|c|=|3x+4y|\leq\sqrt{(3^2+4^2)(x^2+y^2)}=\sqrt{25\cdot4}=10,$$ which says that $3x+4y=10$ is a tangent to the circle,

which is parallel to the line $3x+4y=12$ and nearest to this line.

Id est, the needed distance it's: $$\frac{|12-10|}{\sqrt{3^2+4^2}}=\frac{2}{5}.$$

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