Question

If the line \[ 3x-4y=1 \] touches the circle \[ (x-1)^2+(y+2)^2=4 \] at \((\alpha,\beta)\), the values of \(\alpha\) and \(\beta\) are

• A. \(\alpha=\dfrac{1}{5},\ \beta=-\dfrac{1}{10}\)
• B. \(\alpha=-\dfrac{1}{5},\ \beta=-\dfrac{2}{5}\)
• C. \(\alpha=-\dfrac{2}{5},\ \beta=-\dfrac{11}{20}\)
• D. \(\alpha=\dfrac{2}{5},\ \beta=\dfrac{1}{20}\)

Hint:

The point of contact of a tangent from a circle is obtained by finding the foot of the perpendicular from the centre to the tangent line.

Answer 1

The Correct Option is B

Step 1: Identify the centre and radius of the circle.
Given, \[ (x-1)^2+(y+2)^2=4 \] Comparing with \[ (x-h)^2+(y-k)^2=r^2, \] we get Centre: \[ C=(1,-2) \] Radius: \[ r=2 \]

Step 2: Write the equation of the tangent line.
The given line is \[ 3x-4y=1 \] or \[ 3x-4y-1=0 \]

Step 3: Use the fact that radius is perpendicular to tangent.
The point of contact \((\alpha,\beta)\) is the foot of the perpendicular drawn from the centre \[ (1,-2) \] to the line \[ 3x-4y-1=0 \]

Step 4: Use the foot of perpendicular formula.
For the line \[ ax+by+c=0, \] the foot of perpendicular from \((x_1,y_1)\) is \[ \left( x_1-\frac{a(ax_1+by_1+c)}{a^2+b^2}, \ y_1-\frac{b(ax_1+by_1+c)}{a^2+b^2} \right) \] Here, \[ a=3,\quad b=-4,\quad c=-1 \] and \[ (x_1,y_1)=(1,-2) \]

Step 5: Compute the required quantity.
\[ ax_1+by_1+c = 3(1)+(-4)(-2)-1 \] \[ =3+8-1 \] \[ =10 \] Also, \[ a^2+b^2=3^2+(-4)^2 \] \[ =9+16 \] \[ =25 \]

Step 6: Find \(\alpha\) and \(\beta\).
\[ \alpha = 1-\frac{3(10)}{25} \] \[ = 1-\frac{30}{25} \] \[ = 1-\frac{6}{5} \] \[ = -\frac{1}{5} \] Now, \[ \beta = -2-\frac{(-4)(10)}{25} \] \[ = -2+\frac{40}{25} \] \[ = -2+\frac{8}{5} \] \[ = -\frac{10}{5}+\frac{8}{5} \] \[ = -\frac{2}{5} \]

Step 7: Final conclusion.
Therefore, \[ \boxed{\alpha=-\frac{1}{5},\ \beta=-\frac{2}{5}} \]