I can kill this question in 30 Seconds, can uou?

Geometry Level 3

A circle S = 0 S=0 passes through the point of intersection of two circles: S 1 : x 2 + y 2 − 3 x + 4 y + 5 = 0 S 2 : x 2 + y 2 − 4 x + 3 y + 5 = 0 \displaystyle{{ S }_{ 1 }:\quad { x }^{ 2 }+{ y }^{ 2 }-3x+4y+5=0\\ { S }_{ 2 }:\quad { x }^{ 2 }+{ y }^{ 2 }-4x+3y+5=0}

Circle S = 0 S=0 also cuts the circle S 3 : x 2 + y 2 = 4 { S }_{ 3 }:\quad { x }^{ 2 }+{ y }^{ 2 }=4 orthogonally.

Compute the value of length of tangent from origin to the circle S = 0 S=0 ?

I kill this question orally in a text book for IIT JEE, So I just wanted share this with our community! Hope you may also solved it in this time constrained.


The answer is 2.

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Deepanshu Gupta by Deepanshu Gupta , 25, India

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4 solutions

Krishna Sharma
Dec 30, 2014

First 2 equation are not needed at all as we need to find distance from origin which equals S 1 = c \sqrt{S_1} = \sqrt{c} where 'c' is constant term of that circle

From third equation we will get

2 g 1 . g 2 + 2 f 1 . f 2 = c 1 + c 2 2g_1.g_2 + 2f_1.f_2 = c_1 + c_2

g 1 , f 1 = 0 g_1, f_1 = 0 (circle centred at origin)

Therefore

c 1 = 4 c_1 = 4

Length of tangent = c = 2 \sqrt{c} = 2

5 seconds dot xD

6 Helpful 0 Interesting 0 Brilliant 0 Confused

Exactly :) , I think You Have Supersonic Speed ⌣ ¨ \ddot\smile (5 seconds , great )

Deepanshu Gupta - 6 years, 5 months ago

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Not in 30 sec , so gave up

U Z - 6 years, 5 months ago

It is not clearly mentioned that whehter the circle passes through both points of intersection of S 1 {S}_{1} and S 2 {S}_{2} or only thorough one point of intersection.

Ronak Agarwal - 6 years, 3 months ago

Took 15 seconds -_-

Hats off for 5!

Harsh Shrivastava - 4 years, 2 months ago
Âmåñ Patel
Mar 2, 2015

Took a bit more time aproxx 1 min...

0 Helpful 0 Interesting 0 Brilliant 0 Confused
Fox To-ong
Feb 3, 2015

first we have to solve for the radical axis which can be found by subtracting the equations of two intersecting circles. then we could get y = -x in solving for the length of tangent use pythagorean which is r^2 = x^2 + y^2 where r = the tangent line substitute y = -x to x^2 + y^2 = 4 we could get sqroot of 2 for x and negative for y. then r = 2

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Rudresh Tomar
Dec 31, 2014

10 sec max!

0 Helpful 0 Interesting 0 Brilliant 0 Confused

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