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

Concept: The parametric coordinates of any point on a circle with center \( (h,k) \) and radius \( r \) at angle \( \theta \) are: \[ x = h + r\cos\theta, \quad y = k + r\sin\theta \] Once the point of tangency is found, we write the equation of the tangent line using \( T = 0 \) and compute its distance to the origin.

Step 1: Finding center, radius, and point of tangency.
For the circle \( x^2+y^2-4x-4y+6=0 \):

• Center \( C = (2, 2) \)

• Radius \( r = \sqrt{(-2)^2 + (-2)^2 - 6} = \sqrt{4+4-6} = \sqrt{2} \)
At parametric angle \( \theta = \frac{\pi}{4} \): \[ x = 2 + \sqrt{2}\cos\left(\frac{\pi}{4}\right) = 2 + \sqrt{2}\left(\frac{1}{\sqrt{2}}\right) = 3 \] \[ y = 2 + \sqrt{2}\sin\left(\frac{\pi}{4}\right) = 2 + \sqrt{2}\left(\frac{1}{\sqrt{2}}\right) = 3 \] So the point of contact is \( P(3, 3) \).

Step 2: Writing the equation of the tangent line at (3, 3).
Using the formula \( xx_1 + yy_1 + g(x+x_1) + f(y+y_1) + c = 0 \): \[ 3x + 3y - 2(x + 3) - 2(y + 3) + 6 = 0 \] \[ 3x + 3y - 2x - 6 - 2y - 6 + 6 = 0 \implies x + y - 6 = 0 \]

Step 3: Calculating the perpendicular distance from the origin.
Distance from \( (0,0) \) to \( x + y - 6 = 0 \): \[ d = \frac{|0 + 0 - 6|}{\sqrt{1^2 + 1^2}} = \frac{6}{\sqrt{2}} = 3\sqrt{2} \] This matches Option (B) perfectly.