# shortest distance of $x^2 + y^2 = 4$ and $3x + 4y = 12$

$$x^2 + y^2 = 4$$ 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?

• The line intyersects the circle at $(\frac65,\frac85)$ for me. – Bernard Jul 26 '19 at 17:15
• you still have the point where the line from the origin (with slope $\frac43$) intersects $3x+4y=12$ wrong – J. W. Tanner Jul 26 '19 at 17:20

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 $$\frac{|b|}{\sqrt{1+m^2}}$$

• yes. why i didn't notice it.. – Lifeforbetter Jul 26 '19 at 17:38
• We are all only human beings ... – Michael Hoppe Jul 26 '19 at 17:38

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.

• yes, it doesn't make sense. thank you.. – Lifeforbetter Jul 26 '19 at 17:26

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

• So answer must be $\frac{2}{5}$ I guess? – user572457 Jul 26 '19 at 17:24
• ok, thank you.. but not really got it.. – Lifeforbetter Jul 26 '19 at 17:43
• @Lifeforbetter for any line $ax+by+c=0$ and any point (h,k), the perpendicular of the line from (h,k) is $\frac{|ah+bk+c|}{\sqrt{a^2+b^2}}$ – user572457 Jul 26 '19 at 17:56
• @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. – user572457 Jul 26 '19 at 17:58
• @LifeForBetter $$-10-12\le 6\cos t+8\sin t-12\le10-12$$ right? – lab bhattacharjee Jul 26 '19 at 18:21

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.

• yes. thank you.. – Lifeforbetter Jul 26 '19 at 17:26

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

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: $$\frac{|12-10|}{\sqrt{3^2+4^2}}=\frac{2}{5}.$$