Question

What is the equation of the tangent to the circle ${x^2} + {y^2} + 4x + 6y - 12 = 0$ at (1,1)?
$\left( {\text{a}} \right){\text{ }}3x + 4y = 7 \\\ \left( {\text{b}} \right){\text{ }}3x - 4y = 7 \\\ \left( {\text{c}} \right){\text{ }} - 3x + 4y = 7 \\\ \left( {\text{d}} \right){\text{ }} - 3x - 4y = 7 \\\$

Answer 1

Hint: Here, we will proceed by comparing the given equation of the circle with the general equation of any circle i.e., ${x^2} + {y^2} + 2gx + 2fy + c = 0$. Then, we will use that the formula for the equation of the tangent to any circle at point $\left( {{x_1},{y_1}} \right)$ is given by $x{x_1} + y{y_1} + g\left( {x + {x_1}} \right) + f\left( {y + {y_1}} \right) + c = 0$.

Complete step-by-step answer:
The equation of the given circle is

$${x^2} + {y^2} + 4x + 6y - 12 = 0 \\\ \Rightarrow {x^2} + {y^2} + 4x + 6y + \left( { - 12} \right) = 0{\text{ }} \to (1{\text{)}} \\\$$

As we know that the equation of the tangent to the circle ${x^2} + {y^2} + 2gx + 2fy + c = 0{\text{ }} \to {\text{(2)}}$ at point $\left( {{x_1},{y_1}} \right)$ is given by
$x{x_1} + y{y_1} + g\left( {x + {x_1}} \right) + f\left( {y + {y_1}} \right) + c = 0{\text{ }} \to {\text{(3)}}$
By comparing equation (1) with equation (2), we get
$2g = 4 \\\ \Rightarrow g = 2 \\\$
$2f = 6 \\\ \Rightarrow f = 3 \\\$
c = -12
Since, we have to find the equation of the tangent to the given circle at the point (1,1) which means ${x_1} = 1$ and ${y_1} = 1$.
By substituting g = 2, f = 3, c = -12, ${x_1} = 1$ and ${y_1} = 1$ in equation (3), we get
$\Rightarrow x\left( 1 \right) + y\left( 1 \right) + 2\left( {x + 1} \right) + 3\left( {y + 1} \right) + \left( { - 12} \right) = 0 \\\ \Rightarrow x + y + 2x + 2 + 3y + 3 - 12 = 0 \\\ \Rightarrow 3x + 4y - 7 = 0 \\\ \Rightarrow 3x + 4y = 7 \\\$
Therefore, the required equation of the tangent to the given circle is $3x + 4y = 7$.
Hence, option (a) is correct.

Note- Tangent to any circle is a line which touches that circle at only one point. The perpendicular distance of the center of the circle from the tangent line to that circle is always equal to the radius of the circle. Also, the slope of the tangent to any circle is equal to $\dfrac{{dy}}{{dx}}$ whereas the slope of the normal to that circle is equal to $- \left( {\dfrac{{dx}}{{dy}}} \right)$.