This question is wrong and meaningless and has nothing to do with your courses and entrance exam...but was seems to be interesting..................

As we know that two perpendicular lines has product of slope equal to -1.

Like suppose  y=m1x and y=m2 x is two perpendicular line...........

Then  m1.m2 = -1.

...............then y=0  and  x=0    these two line are also perpendicular......but here we have not this condition.....of m1.m2=-1...............here we have  m1.m2 =0..............

So is y=0 and x=0 are not perpendicular.............or m1.m2 = -1 is wrong............or still  you can prove that.....slope of y=0 and x=0 has product = -1.

 No body else is interested in question.!!!!!!!!!!!!!!!!!!!

very good question

lets start from fundamental  where did you drive that m1 m2=-1

lets start proof

lets  y=m1x+c1,y=m2x+c2 are two lines

to find angle between them say theta

tan theta =tan(theta1-theta2)= (tan (theta1) -tan (theta2))/(1+tan (theta1)tan (theta2))

for lines to be perpendiculars

tan theta  should be infinity

means (1+tan (theta1)tan (theta2)) =0

1+m1m2=0

proved

now coming back to your question

y = c and x =b are two lines

m1 = 0 and m2 = infinity in the very first equation you wil get the result

 E: x²/a²+y²/b²=1 and C: x²+y²+2gx+2fy+c=0

then equation E+lemda C=0 represent a curve which passess through the comman point of the ellipse and circle..

Qus-The Center of circle C which passess throuch the three points on E whose eccentric angles are Alpha,Beta,Gamma.....