If the circle x^2 + y^2 +4x+2y-4=0 is reflected in a mirror to become a circle x^ + y^2 +6x+4y+4=0 the mirror is lying on the straight line?

Question

Meet · Answer

Here, let the line be ax+by+c=0Now, ((x-x1)/a)=((y-y1) /b) where x, y are coordinates of centre of new circle and x1,y1 are coordinates of centre of old circle. Solving this we get a=bNow,on solving[(x-x1)/a]=[-2(ax1+by1+c)]/[a^2+b^2]we get -4a=cAnd on substituting a=b=-c/4 in line equation we get linex+y-4=0

Meet · Answer

Here, let the line be ax+by+c=0Now, ((x-x1)/a)=((y-y1) /b) where x, y are coordinates of centre of new circle and x1,y1 are coordinates of centre of old circle. Solving this we get a=bNow,on solving[(x-x1)/a]=[-2(ax1+by1+c)]/[a^2+b^2]we get -4a=cAnd on substituting a=b=-c/4 in line equation we get linex+y-4=0There is a mistake we get 4a=cSo we get line x+y+4=0

Thank you for registering.

Register Now and Win Upto 25% Scholorship for a Full Academic Year !

Enter Your Details

Registration done!

If the circle x^2 + y^2 +4x+2y-4=0 is reflected in a mirror to become a circle x^ + y^2 +6x+4y+4=0 the mirror is lying on the straight line?

If the circle x^2 + y^2 +4x+2y-4=0 is reflected in a mirror to become a circle x^ + y^2 +6x+4y+4=0 the mirror is lying on the straight line?

2 Answers

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

Other Related Questions on Analytical Geometry

ASK QUESTION

Answer and earn gift vouchers

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Complete Self Study Package designed by Industry Leading Experts

Live 1-1 coding classes to unleash the Creator in your Child

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Register Now & Win Upto 25% Scholorship