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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?
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
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
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