The tangent to the circle centred at (0, 0) and with radius = 1 at point \(\left(\displaystyle\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)\) on it is given by
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(x+y=\sqrt{2}\)
The circle with centre (0, 0) and radius 1 is
\(x^2+y^2=1\)
Differentiating implicitly with respect to x:
\( \displaystyle 2x+2y\,\frac{dy}{dx}=0 \)
\( \displaystyle \frac{dy}{dx}=-\frac{x}{y} \)
At \(\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)\) the slope of the tangent is \(-1\). The tangent is
\( \displaystyle y-\frac{1}{\sqrt{2}}=-\left(x-\frac{1}{\sqrt{2}}\right) \)
\( \displaystyle x+y=\frac{2}{\sqrt{2}} \)
\(x+y=\sqrt{2}\)
Shortcut: The tangent to \(x^2+y^2=a^2\) at the point \((x_1, y_1)\) on it is \(xx_1+yy_1=a^2\). Here
\( \displaystyle \frac{x}{\sqrt{2}}+\frac{y}{\sqrt{2}}=1 \)
which is \(x+y=\sqrt{2}\).
Note: This topic was not in the CUET 2026 syllabus.