Given:

The curve x² + y² = 13 and the line 2x + 3y = 7

x² + y² = 13

Differentiating the above w.r.t x

⇒ 2x^{2 – 1} + 2y^{2 – 1} \(\frac{dy}{dx}\)= 0

⇒ 2x + 2y\(\frac{dy}{dx}\) = 0

⇒ 2(x + y\(\frac{dy}{dx}\) ) = 0

⇒ (x + y\(\frac{dy}{dx}\) ) = 0

⇒ y\(\frac{dy}{dx}\)= – x

⇒ \(\frac{dy}{dx}\) = \(\frac{-x}{y}\)...(1)

Since, line is 2x + 3y = 7

⇒ 3y = – 2x + 7

⇒ y = \(\frac{-2x+7}{3}\)

⇒ y = \(\frac{-2x}{3}+\frac{7}{3}\)

\(\therefore\) The equation of a straight line is y = mx + c, where m is the The Slope of the line.

Thus, the The Slope of the line is \(\frac{-2}{3}\)...(2)

Since, tangent is parallel to the line,

\(\therefore\) the The Slope of the tangent = The Slope of the normal

\(\frac{-x}{y}\) = \(\frac{-2}{3}\)

⇒ – x = \(\frac{-2y}{3}\)

⇒ x = \(\frac{2y}{3}\)

Substituting x = \(\frac{2y}{3}\) in x² + y² = 13,

⇒ ( \(\frac{2y}{3}\) )² + y² = 13

⇒ ( \(\frac{4y^2}{3}\) ) + y² = 13

⇒ y²( \(\frac{4}{9}+1\) ) = 13

⇒ y²( \(\frac{13}{9}\) ) = 13

⇒ y²( \(\frac{1}{9}\) ) = 1

⇒ y² = 9

⇒ y = \(\pm\)3

Substituting y = \(\pm\)3 in x = \(\frac{2y}{3}\),we get,

x = \(\frac{2x(\pm3)}{3}\)

x = \(\pm\)2

Thus, the required point is (2, 3) & ( – 2, – 3)