Question

Consider the circle $x^{2}+y^{2}=1$. The point on the circle which is nearest from the point (1,1), is

• A. $\left(-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)$
• B. $\left(\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\right)$
• C. $\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)$
• D. $\left(-\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\right)$
• E. $(0,1)$

Hint:

Logic Tip: You can often skip the math entirely by visualizing the geometry! The point $(1,1)$ is perfectly diagonal in the top-right quadrant. The closest edge of a circle centered at $(0,0)$ must logically be exactly in the middle of that top-right quadrant (where $x$ and $y$ are equal and positive). Option C is the only one that fits!

Answer 1

The Correct Option is C

Concept:
Coordinate Geometry - Shortest Distance to a Circle.
The shortest (or longest) distance from any external point to a circle always lies along the normal line that passes through the center of the circle.
Step 1: Identify the properties of the given circle.
The equation is $x^2 + y^2 = 1$.
This is a standard circle with its center at the origin $C(0,0)$ and a radius $r = 1$.
Step 2: Find the equation of the normal line.
The normal line must pass through the center $C(0,0)$ and the external point $P(1,1)$.
Calculate the slope $m$ of this line: $$ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{1 - 0}{1 - 0} = 1 $$ The equation of a line passing through the origin is $y = mx$. Thus, the line is: $$ y = x $$
Step 3: Find the intersection points with the circle.
Substitute $y = x$ into the circle's equation to find where this normal line cuts the boundary of the circle: $$ x^2 + (x)^2 = 1 $$ $$ 2x^2 = 1 $$ $$ x^2 = \frac{1}{2} $$ Take the square root to solve for $x$: $$ x = \pm\frac{1}{\sqrt{2}} $$
Step 4: Determine the corresponding y-coordinates.
Since the intersection points lie on the line $y = x$, the $y$-coordinates will be identical to the $x$-coordinates. This gives two intersection points: $$ A\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right) \quad \text{and} \quad B\left(-\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}\right) $$
Step 5: Select the nearest point.
The point $P(1,1)$ lies strictly in the first quadrant (both coordinates are positive).
To minimize the distance, the nearest point on the circle must also lie in the first quadrant facing $P$.
Therefore, we select point $A$ where both coordinates are positive: $$ \text{Nearest Point} = \left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right) $$