Question

The point (3 , 5) lies ..... the circle with equation x²+y² = 9?

Answer 1

Answer:

Any point on the lines 7x+y+3=0 is Q(t,−3−7t),t∈R.

Now P (h,k) is image of point Q in the line x−y+1=0

Then, 1h−t=−1k−(−3−7t)

         =−1+12(t−(−3−7t)+1)

         =−8t−4

⇒(h,k)≡(−7t−4,t+1)

This point lies on the circle x2+y2=9

⇒(−7t−4)2+(t+1)2=9

⇒50r2+58t+8=0

⇒25r2+19t+4=0

⇒(25t+4)(t+1)=0

⇒t=−4/25,t=1

⇒(h,k)=(−2572,2521) or (3,0).