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