Question

WHO ANSWER THIS ?
A tangent at first quadrant is drawn to the circle\[x^{2}+y^{2}=4\]so that the part intercepted by the tangent line is a minimum. Then find the equation of the tangent line. (The question is to mean that that line of tangency must possess a minimum magnitude in length b/n the two intersection points on the axes).

Answer 1

that tangent line is perpendecular to radius of circle like this

Answer 2

triangles OAB and OBC are right triangles so \[AB=r \tan(\frac{\pi}{2}-\alpha)\]\[BC=r \tan(\alpha)\]\[r=\sqrt{2}\]

Answer 3

so u need to minimize \[f(\alpha)=r(\tan \alpha+\cot \alpha)\]

Answer 4

and note that\[0<\alpha<\frac{\pi}{2}\]

Answer 5

can u do that?

Answer 6

r=2

Answer 7

@sauravshakya @mukushla
I ask you to find the equation of the tangent line, not the length of the line b/n the points intercept.
Will you try again?

Answer 8

Here is the answer…..
Let the tangent line be y = mx+b or y-mx-b = 0; so that the y and x – intercepts of the line are (0,b) and (-b/m,0). Thus the length of l is given as:\[l=\sqrt{b^{2}+(\frac{-b}{m})^{2}}.............(1)\]Now lets express ‘ b ‘ in terms of ‘ m ‘,or vice versa. To do that we need to find the distance from the origin to the line:\[d=\frac{ \left| Ax+By+C \right| }{\sqrt{A^{2}+B^{2}} }\]\[=2=\frac{ \left| (-m)(0)^{2}+(1)(0)^{2}-b \right| }{ \sqrt{m^{2}+1} }\]\[=4=\frac{b^{2}}{m^2+1}\]\[=b^{2}=4(m^{2}+1).......(2)\]Inserting equation 2 in 1, we get \[l=\sqrt{4(m^2+1)+\frac{4(m^{2}+1)}{m^{2}}}\]\[l=\frac{2(m^{2}+1)}{m}\]Thus to be the length of l minimum this condition must be satisfied \[\frac{ d }{ dm }l=0=\frac{ 2m^{2}-2 }{m^{2} }\] \[=m=\pm1\]But m = -1 is the only answer since the slope of the line is negative. And \[b =\pm2\sqrt{2}....due to equation (2)\]But \[b =2\sqrt{2}\]is the only answer. Therefore the line would be\[ y=mx+b{\rightarrow}y=-x+ 2\sqrt{2}\]

Answer 9

closing the question......

Answer 10

WELL YOUR METHOD IS VERY CORRECT....... and so is @mukushla 's method.
The minimum length of AC is 4 when alpha=45 Now, this gives