Question

Let's solve JEE questions!

Answer 1

A tangent PT is drawn to the circle \(x^2 + y^2 = 4\) at the point \(P(\sqrt 3, 1)\). A straight line \(L\), perpendicular to PT is a tangent to the circle \((x-3)^2 + y^2= 1\).

Answer 2

idk

Answer 3

Q1: What are the possible equations of \(L\)?

Answer 4

Q2: What are the common tangents to the circles?

Answer 5

uh what????

Answer 6

Let's do this! The equation of the tangent at \(P\) is \(\sqrt{3}x + y=4\). Now the equation of the perpendicular is of the form \(\sqrt{3}y - x = k\). When will this be tangent to the circle? Simply when the distance of the line from the center \( (3,0)\) is \(1\).

Answer 7

Im sorry bro but I honestly dont understand these what grade are you in?

Answer 8

\[|-3 - k|/2 = 1 \Rightarrow |3+k| = 2 \Rightarrow k = -1, k = -5\]

Answer 9

\[x - \sqrt{3}y = 1\]\[x - \sqrt{3}y = 5\]We obtained the answer to this question. Woo!

Answer 10

Ah, good question. Why not use differentiation?

Answer 11

There's actually an easier formula for this. Whenever you're given ANY second degree equation of the form\[ax^2 + by^2 + 2hxy + 2gx + 2fy + c = 0\]and a point \((x_1, y_1)\) at which you want to find the tangent's equation, what you do is this:

Replace \(x^2\) by \(xx_1\)
Replace \(y^2\) by \(yy_1\)
Replace \(2x\) by \(x + x_1\)
Replace \(2y\) by \(y+y_1\)

Answer 12

So that's quite a useful trick which I used.

Answer 13

yes you could say that it is parallel to the radius so you wouldn't even need the equation of the tangent.

Answer 14

Alright, any ideas for the second question then?

Answer 15

But I'm the asker. lol

Answer 16

I am.

Answer 17

What are you talking about?

Answer 18

But I'm the one who asked this question, I promise. Haha.

Answer 19

I'm having this problem with Chrome where after I click the post button, everything freezes.

Answer 20

parth's gone mad......:-)

for perp's to anything use the gradient and it's vector.

Answer 21

Not sure if you're still wondering about question 2, I believe it would just be the tangent at the point where the circles intersect.

Answer 22

Err, but that account is really mine. How do I fix this problem?

@Alphabet_Sam There are more of them.

Answer 23

Ahh I see.

Answer 24

Hint: use similarity to find the point where those two tangents meet

Answer 25

ok this has become boring

Answer 26

@baru say something.

Answer 27

indeed I will.
@studying24x7 REVEAL YOUR TRUE IDENTITY!!! your comfort level with latex is too high for you to be new.

Answer 28

lol that is me.

Answer 29

:O

Answer 30

The axis of a parabola is along the line \(y = x\). The distance of its vertex from the origin is \(\sqrt{2}\) and the distance of its focus is \(2\sqrt 2\). If the vertex and the focus both lie in the first quadrant, the equation of the parabola is...

Answer 31

Directrix: \(x+y=0\)
Focus \((2,2)\)

Equation:\[|x+y|/\sqrt{2}=\sqrt{(x-2)^2+(y-2)^2}\]