Question

The perpendicular distance from origin to the tangent drawn at the point \( P(\frac{\pi}{4}) \) to the circle \( x^{2}+y^{2}-4x-4y+6=0 \) is

• A. 4
• B. \( 3\sqrt{2} \)
• C. 6
• D. \( 5\sqrt{2} \)

Hint:

The tangent line is perpendicular to the normal line connecting the center and the contact point. The vector from the center \( (2,2) \) to \( (3,3) \) is \( (1,1) \), confirming that the tangent line slope must be \( -1 \) via a quick mental check.

Answer 1

The Correct Option is B

Step 1: Find the centre and radius.
The circle $x^2+y^2-4x-4y+6=0$ has centre $(2,2)$ and radius $r=\sqrt{(-2)^2+(-2)^2-6}=\sqrt{4+4-6}=\sqrt{2}$.
Step 2: Use the parametric point.
A point on the circle at angle $\theta$ is $(2+r\cos\theta,\ 2+r\sin\theta)$. At $\theta=\frac{\pi}{4}$, $\cos\theta=\sin\theta=\frac{1}{\sqrt{2}}$.
Step 3: Find the point of contact $P$.
\[ x=2+\sqrt{2}\cdot\tfrac{1}{\sqrt{2}}=3,\qquad y=2+\sqrt{2}\cdot\tfrac{1}{\sqrt{2}}=3 \] So $P=(3,3)$.
Step 4: Write the tangent at $P$.
The tangent at $(x_1,y_1)$ uses $xx_1+yy_1+g(x+x_1)+f(y+y_1)+c=0$ with $g=-2$, $f=-2$, $c=6$: \[ 3x+3y-2(x+3)-2(y+3)+6=0 \]
Step 5: Simplify the tangent line.
\[ 3x+3y-2x-6-2y-6+6=0\implies x+y-6=0 \]
Step 6: Distance from origin to this tangent.
\[ d=\frac{|0+0-6|}{\sqrt{1^2+1^2}}=\frac{6}{\sqrt{2}}=3\sqrt{2} \] \[ \boxed{3\sqrt{2}} \]