Question

The equation of a tangent to the circle \[ x^2+y^2=1, \] which is perpendicular to the line \[ y=mx+1, \] is

• A. \(x+my-\sqrt{1+m^2}=0\)
• B. \(mx+y-\sqrt{1+m^2}=0\)
• C. \(x+my+\sqrt{1+m^2}=0\)
• D. \(mx+y+\sqrt{1+m^2}=0\)

Hint:

For the circle \[ x^2+y^2=a^2, \] the tangent with slope \(m\) is \[ y=mx\pm a\sqrt{1+m^2}. \]

Answer 1

The Correct Option is A

Step 1: Find the slope of the given line.
The line is \[ y=mx+1 \] Hence, its slope is \[ m \]

Step 2: Determine the slope of the perpendicular tangent.
If two lines are perpendicular, then the product of their slopes is \[ -1 \] Therefore, the slope of the tangent is \[ -\frac{1}{m} \]

Step 3: Write the general equation of the tangent.
A line with slope \[ -\frac{1}{m} \] can be written as \[ x+my+c=0 \]

Step 4: Use the tangent condition for the circle.
The circle is \[ x^2+y^2=1 \] Its centre is \[ (0,0) \] and radius is \[ 1 \] The distance of the tangent from the centre must equal the radius.

Step 5: Apply the distance formula.
Distance of \[ x+my+c=0 \] from \[ (0,0) \] is \[ \frac{|c|}{\sqrt{1+m^2}} \] Since this equals the radius \(1\), \[ \frac{|c|}{\sqrt{1+m^2}}=1 \] Therefore, \[ |c|=\sqrt{1+m^2} \] \[ c=\pm\sqrt{1+m^2} \]

Step 6: Write the tangent equations.
Hence, the tangent equations are \[ x+my\pm\sqrt{1+m^2}=0 \] Among the given options, \[ x+my-\sqrt{1+m^2}=0 \] matches.

Step 7: Final conclusion.
Therefore, the required tangent is \[ \boxed{x+my-\sqrt{1+m^2}=0} \]