Question Statement

The point $P (- 11, - 10)$ lies outside the circle $x^{2} + y^{2} + 6 x + 8 y + 5 = 0$ . Find the equations of the tangents to the circle drawn from point $P$ and find the points of contact.

Background and Explanation

To solve this problem, we use the property that a tangent to a circle is perpendicular to the radius at the point of contact. By representing the point of contact as $(x_{1}, y_{1})$ , we can use the product of slopes and the fact that the point lies on the circle to solve for the coordinates. Finally, the equations of the tangents are found using the two-point form for a line.

Solution

Finding the Center of the Circle

The given equation of the circle is: $x^{2} + y^{2} + 6 x + 8 y + 5 = 0$ Comparing this to the general form $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ , we find the center $C (g, f)$ : $C (\frac{6}{- 2}, \frac{8}{- 2}) = (- 3, - 4)$

Establishing the Geometric Relationship

Let $A (x_{1}, y_{1})$ and $B$ be the points of contact where the tangents from $P (- 11, - 10)$ touch the circle.

Since the radius $\overline{A C}$ is perpendicular to the tangent $\overline{A P}$ , the product of their slopes is $- 1$ : $(Slope of \overline{A C}) \times (Slope of \overline{A P}) = - 1$

Substituting the coordinates $C (- 3, - 4)$ , $P (- 11, - 10)$ , and $A (x_{1}, y_{1})$ : $(\frac{y _{1} - ( - 4 )}{x _{1} - ( - 3 )}) (\frac{y _{1} - ( - 10 )}{x _{1} - ( - 11 )}) = - 1$ $\Rightarrow (\frac{y _{1} + 4}{x _{1} + 3}) (\frac{y _{1} + 10}{x _{1} + 11}) = - 1$ $\frac{y _{1}^{2} + 10 y _{1} + 4 y _{1} + 40}{x _{1}^{2} + 11 x _{1} + 3 x _{1} + 33} = - 1$ $\frac{y _{1}^{2} + 14 y _{1} + 40}{x _{1}^{2} + 14 x _{1} + 33} = - 1$ $y_{1}^{2} + 14 y_{1} + 40 = - 1 (x_{1}^{2} + 14 x_{1} + 33)$ $y_{1}^{2} + 14 y_{1} + 40 = - x_{1}^{2} - 14 x_{1} - 33$ $x_{1}^{2} + y_{1}^{2} + 14 x_{1} + 14 y_{1} + 73 = 0 \dots (i)$

Finding the Points of Contact

Since the point $A (x_{1}, y_{1})$ lies on the circle, it must satisfy the circle's equation: $x_{1}^{2} + y_{1}^{2} + 6 x_{1} + 8 y_{1} + 5 = 0 \dots (ii)$

Subtracting equation (ii) from equation (i): $(x_{1}^{2} - x_{1}^{2}) + (y_{1}^{2} - y_{1}^{2}) + (14 x_{1} - 6 x_{1}) + (14 y_{1} - 8 y_{1}) + (73 - 5) = 0$ $8 x_{1} + 6 y_{1} + 68 = 0$ Dividing by 2: $4 x_{1} + 3 y_{1} + 34 = 0 \dots (iii)$

From (iii), we express $y_{1}$ in terms of $x_{1}$ : $3 y_{1} = - 4 x_{1} - 34 \Rightarrow y_{1} = \frac{- 4 x _{1} - 34}{3}$

Substitute this into the circle equation (ii): $x_{1}^{2} + (\frac{- 4 x _{1} - 34}{3})^{2} + 6 x_{1} + 8 (\frac{- 4 x _{1} - 34}{3}) + 5 = 0$ $x_{1}^{2} + \frac{16 x _{1}^{2} + 272 x _{1} + 1156}{9} + 6 x_{1} + \frac{- 32 x _{1} - 272}{3} + 5 = 0$

Multiply the entire equation by 9 to clear the denominators: $9 x_{1}^{2} + 16 x_{1}^{2} + 272 x_{1} + 1156 + 54 x_{1} - 96 x_{1} - 816 + 45 = 0$ $25 x_{1}^{2} + 230 x_{1} + 385 = 0$ Divide by 5: $5 x_{1}^{2} + 46 x_{1} + 77 = 0$ Factor the quadratic: $5 x_{1}^{2} + 35 x_{1} + 11 x_{1} + 77 = 0$ $5 x_{1} (x_{1} + 7) + 11 (x_{1} + 7) = 0$ $(x_{1} + 7) (5 x_{1} + 11) = 0$

This gives two possible values for $x_{1}$ :

$x_{1} = - 7$
$x_{1} = - \frac{11}{5}$

Case 1: $x_{1} = - 7$ Substitute into (iii): $4 (- 7) + 3 y_{1} + 34 = 0 \Rightarrow - 28 + 3 y_{1} + 34 = 0$ $3 y_{1} = - 6 \Rightarrow y_{1} = - 2$ Point $A = (- 7, - 2)$ .

Case 2: $x_{1} = - \frac{11}{5}$ Substitute into (iii): $4 (- \frac{11}{5}) + 3 y_{1} + 34 = 0 \Rightarrow - \frac{44}{5} + 3 y_{1} + \frac{170}{5} = 0$ $3 y_{1} = - \frac{126}{5} \Rightarrow y_{1} = - \frac{42}{5}$ Point $B = (- \frac{11}{5}, - \frac{42}{5})$ .

Finding the Equations of the Tangents

We use the two-point formula: $\frac{y - y _{P}}{y _{A} - y _{P}} = \frac{x - x _{P}}{x _{A} - x _{P}}$

Tangent through $P (- 11, - 10)$ and $A (- 7, - 2)$ : $\frac{y - ( - 10 )}{- 2 - ( - 10 )} = \frac{x - ( - 11 )}{- 7 - ( - 11 )}$ $\frac{y + 10}{8} = \frac{x + 11}{4}$ $\frac{y + 10}{2} = x + 11$ $y + 10 = 2 x + 22$ $2 x - y + 12 = 0$

Tangent through $P (- 11, - 10)$ and $B (- \frac{11}{5}, - \frac{42}{5})$ : $\frac{y - ( - 10 )}{- \frac{42}{5} - ( - 10 )} = \frac{x - ( - 11 )}{- \frac{11}{5} - ( - 11 )}$ $\frac{y + 10}{\frac{8}{5}} = \frac{x + 11}{\frac{44}{5}}$ $\frac{y + 10}{8} = \frac{x + 11}{44}$ $\frac{y + 10}{2} = \frac{x + 11}{11}$ $11 (y + 10) = 2 (x + 11)$ $11 y + 110 = 2 x + 22$ $2 x - 11 y - 88 = 0$

Key Formulas or Methods Used

Center of a Circle: For $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ , Center is $(- g, - f)$ .
Slope Formula: $m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}$ .
Perpendicular Lines: $m_{1} \cdot m_{2} = - 1$ .
Two-Point Form of a Line: $\frac{y - y _{1}}{y _{2} - y _{1}} = \frac{x - x _{1}}{x _{2} - x _{1}}$ .

Summary of Steps

Identify the center of the circle $C (- 3, - 4)$ .
Set up the perpendicularity condition between the radius $C A$ and the tangent $P A$ .
Combine the resulting equation with the circle's equation to find the points of contact.
Solve the quadratic equation to find the $x$ -coordinates of the contact points $A$ and $B$ .
Find the corresponding $y$ -coordinates using the linear relation derived from the two equations.
Use the coordinates of $P$ and the contact points to write the equations of the two tangents using the two-point formula.

Question Statement

The point $P (- 11, - 10)$ lies outside the circle $x^{2} + y^{2} + 6 x + 8 y + 5 = 0$ . Find the equations of the tangents to the circle drawn from point $P$ and find the points of contact.

Background and Explanation

Solution

Finding the Center of the Circle

Establishing the Geometric Relationship

Let $A (x_{1}, y_{1})$ and $B$ be the points of contact where the tangents from $P (- 11, - 10)$ touch the circle.

Since the radius $\overline{A C}$ is perpendicular to the tangent $\overline{A P}$ , the product of their slopes is $- 1$ : $(Slope of \overline{A C}) \times (Slope of \overline{A P}) = - 1$

Finding the Points of Contact

Since the point $A (x_{1}, y_{1})$ lies on the circle, it must satisfy the circle's equation: $x_{1}^{2} + y_{1}^{2} + 6 x_{1} + 8 y_{1} + 5 = 0 \dots (ii)$

From (iii), we express $y_{1}$ in terms of $x_{1}$ : $3 y_{1} = - 4 x_{1} - 34 \Rightarrow y_{1} = \frac{- 4 x _{1} - 34}{3}$

This gives two possible values for $x_{1}$ :

$x_{1} = - 7$
$x_{1} = - \frac{11}{5}$

Case 1: $x_{1} = - 7$ Substitute into (iii): $4 (- 7) + 3 y_{1} + 34 = 0 \Rightarrow - 28 + 3 y_{1} + 34 = 0$ $3 y_{1} = - 6 \Rightarrow y_{1} = - 2$ Point $A = (- 7, - 2)$ .

Finding the Equations of the Tangents

We use the two-point formula: $\frac{y - y _{P}}{y _{A} - y _{P}} = \frac{x - x _{P}}{x _{A} - x _{P}}$

Key Formulas or Methods Used

Center of a Circle: For $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ , Center is $(- g, - f)$ .
Slope Formula: $m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}$ .
Perpendicular Lines: $m_{1} \cdot m_{2} = - 1$ .
Two-Point Form of a Line: $\frac{y - y _{1}}{y _{2} - y _{1}} = \frac{x - x _{1}}{x _{2} - x _{1}}$ .

Summary of Steps

Identify the center of the circle $C (- 3, - 4)$ .
Set up the perpendicularity condition between the radius $C A$ and the tangent $P A$ .
Combine the resulting equation with the circle's equation to find the points of contact.
Solve the quadratic equation to find the $x$ -coordinates of the contact points $A$ and $B$ .
Find the corresponding $y$ -coordinates using the linear relation derived from the two equations.
Use the coordinates of $P$ and the contact points to write the equations of the two tangents using the two-point formula.

7.3 Q-4

Question Statement

Background and Explanation

Solution

Finding the Center of the Circle

Establishing the Geometric Relationship

Finding the Points of Contact

Finding the Equations of the Tangents

Key Formulas or Methods Used

Summary of Steps

7.3 Q-4

Question Statement

Background and Explanation

Solution

Finding the Center of the Circle

Establishing the Geometric Relationship

Finding the Points of Contact

Finding the Equations of the Tangents

Key Formulas or Methods Used

Summary of Steps