Question

The tangent to the curve $x^2+y^2=25$ parallel to the line 3x-4y=7 exist at the point

• A. (-3, -4)
• B. (3, -4)
• C. (3, 4)
• D. (1,1)

Answer 1

Answer: (A), (B)

(3, -4)

We have,
$x^2+y^2=25\Rightarrow 2x+2y\frac{dy}{dx}=0\Rightarrow \frac{dy}{dx}=-\frac{x}{y}$.
Now, slope of the line 3x -4y =7 is m =$\frac{3}{4}$.
Since the tan gent is parallel to the given line.
$\Rightarrow \frac{dy}{dx}=\frac{3}{4}\Rightarrow \frac{x}{y}=\frac{3}{4}\Rightarrow y=-\frac{4}{3}x$.
From equation (1): $x^2+\frac{16}{9}{x}^2=25\Rightarrow x=\pm 3$.
If x =3, from equation (2), $y=-\frac{4}{3}\left(3\right)=-4$.
If x =-3, from equation (2), $y=-\frac{4}{3}\left(3\right)=-4$.
The point s are (3,-4) and (-3, 4).
Hence (b) is the correct answer.