The tangent to the curve x2+y2=25 parallel to the line 3x-4y=7 exist at the point
(3, -4)
We have,
x2+y2=25⇒2x+2ydydx=0⇒dydx=−xy.
Now, slope of the line 3x -4y =7 is m =34.
Since the tan gent is parallel to the given line.
⇒dydx=34⇒xy=34⇒y=−43x.
From equation (1): x2+169x2=25⇒x=±3.
If x =3, from equation (2), y=−43(3)=−4.
If x =-3, from equation (2), y=−43(3)=−4.
The point s are (3,-4) and (-3, 4).
Hence (b) is the correct answer.