Question Statement

Find the equations of the tangent lines to the circle $(x - 1)^{2} + (y - 2)^{2} = 3^{2}$ that are perpendicular to the line $x + 3 y - 6 = 0$ .

Background and Explanation

This problem requires finding lines that satisfy two conditions: being perpendicular to a given line (which determines the slope) and being tangent to a given circle (which determines the y-intercept). Recall that perpendicular lines have slopes that are negative reciprocals of each other, and a line is tangent to a circle if it intersects the circle at exactly one point.

Solution

Method 1: Algebraic Approach (Using the Discriminant)

Let the equation of the tangent line be $y = m x + c$ .

First, determine the slope $m$ . The given line $x + 3 y - 6 = 0$ can be rewritten as $y = - \frac{1}{3} x + 2$ , so its slope is $- \frac{1}{3}$ . Since the tangent line is perpendicular to this line, its slope is the negative reciprocal: $m = 3$

Thus, the equation of our tangent line becomes: $y = 3 x + c$

To find the value of $c$ , substitute $y = 3 x + c$ into the circle's equation $(x - 1)^{2} + (y - 2)^{2} = 9$ :

$(x - 1)^{2} + (3 x + c - 2)^{2} = 9$

Expand and simplify: $x^{2} - 2 x + 1 + 9 x^{2} + c^{2} + 4 + 6 c x - 12 x - 4 c - 9 = 0$

Combine like terms: $10 x^{2} + (6 c - 14) x + (c^{2} - 4 c - 4) = 0$

Or factoring the coefficient of $x$ : $10 x^{2} + 2 (3 c - 7) x + (c^{2} - 4 c - 4) = 0$

For the line to be tangent to the circle, this quadratic equation must have exactly one solution for $x$ , meaning its discriminant must equal zero: $[2 (3 c - 7)]^{2} - 4 (10) (c^{2} - 4 c - 4) = 0$

Expand the discriminant: $4 (9 c^{2} - 42 c + 49) - 40 (c^{2} - 4 c - 4) = 0$

Divide both sides by 4: $9 c^{2} - 42 c + 49 - 10 c^{2} + 40 c + 40 = 0$

Simplify: $- c^{2} - 2 c + 89 = 0$

Multiply by $- 1$ : $c^{2} + 2 c - 89 = 0$

Solve for $c$ using the quadratic formula: $c = \frac{- 2 \pm ( 2 ) ^{2} - 4 ( 1 ) ( - 89 )}{2 ( 1 )} = \frac{- 2 \pm 4 + 356}{2} = \frac{- 2 \pm 360}{2}$

Simplify $360 = 36 \times 10 = 610$ : $c = \frac{- 2 \pm 6 10}{2} = - 1 \pm 310$

Substituting these values of $c$ back into the line equation $y = 3 x + c$ :

$y = 3 x + (- 1 \pm 310)$

Or equivalently: $y = 3 x - 1 + 310 and y = 3 x - 1 - 310$

Method 2: Geometric Approach (Using Distance Formula)

Again, we establish that the slope of the tangent line must be $m = 3$ (since it is perpendicular to the line with slope $- \frac{1}{3}$ ). Thus, the tangent line has the form: $y = 3 x + c$ or in standard form: $3 x - y + c = 0$

The circle has center $(1, 2)$ and radius $r = 3$ . For a line to be tangent to a circle, the perpendicular distance from the center to the line must equal the radius.

Using the distance formula from point $(x_{0}, y_{0}) = (1, 2)$ to line $3 x - y + c = 0$ : $\frac{∣3 ( 1 ) - ( 2 ) + c ∣}{3 ^{2} + ( - 1 ) ^{2}} = 3$

Simplify: $\frac{∣1 + c ∣}{10} = 3$

Multiply both sides by $10$ : $∣1 + c ∣ = 310$

This gives two cases: $1 + c = 310 or 1 + c = - 310$

Solving for $c$ : $c = - 1 + 310 or c = - 1 - 310$

Substituting back into the standard form equation $3 x - y + c = 0$ :

$3 x - y + (- 1 \pm 310) = 0$

Or equivalently: $3 x - y - 1 + 310 = 0 and 3 x - y - 1 - 310 = 0$

Key Formulas or Methods Used

Perpendicular Slopes: If two lines with slopes $m_{1}$ and $m_{2}$ are perpendicular, then $m_{1} \cdot m_{2} = - 1$ (or $m_{2} = - \frac{1}{m _{1}}$ )
Algebraic Tangency Condition: A line is tangent to a circle if the system of equations yields a quadratic with discriminant $D = b^{2} - 4 a c = 0$
Distance from Point to Line: $d = \frac{∣ A x _{0} + B y _{0} + C ∣}{A ^{2} + B ^{2}}$ for line $A x + B y + C = 0$ and point $(x_{0}, y_{0})$
Geometric Tangency Condition: Distance from circle's center to the line equals the radius ( $d = r$ )

Summary of Steps

Find the tangent slope: Calculate the negative reciprocal of the given line's slope ( $- \frac{1}{3}$ ) to get $m = 3$
Write line equation: Express the tangent line as $y = 3 x + c$ (or $3 x - y + c = 0$ ) with unknown $c$
Apply tangency condition:
- Method 1: Substitute into circle equation, set discriminant of resulting quadratic to zero, and solve for $c$
- Method 2: Set distance from center $(1, 2)$ to line equal to radius $3$ , solve $∣1 + c ∣ = 310$ for $c$
Solve for $c$ : Obtain $c = - 1 \pm 310$
Write final equations: Substitute $c$ values back to get the two tangent lines: $y = 3 x - 1 \pm 310$ or $3 x - y - 1 \pm 310 = 0$

Question Statement

Find the equations of the tangent lines to the circle $(x - 1)^{2} + (y - 2)^{2} = 3^{2}$ that are perpendicular to the line $x + 3 y - 6 = 0$ .

Background and Explanation

Solution

Method 1: Algebraic Approach (Using the Discriminant)

Let the equation of the tangent line be $y = m x + c$ .

Thus, the equation of our tangent line becomes: $y = 3 x + c$

To find the value of $c$ , substitute $y = 3 x + c$ into the circle's equation $(x - 1)^{2} + (y - 2)^{2} = 9$ :

$(x - 1)^{2} + (3 x + c - 2)^{2} = 9$

Expand and simplify: $x^{2} - 2 x + 1 + 9 x^{2} + c^{2} + 4 + 6 c x - 12 x - 4 c - 9 = 0$

Combine like terms: $10 x^{2} + (6 c - 14) x + (c^{2} - 4 c - 4) = 0$

Or factoring the coefficient of $x$ : $10 x^{2} + 2 (3 c - 7) x + (c^{2} - 4 c - 4) = 0$

For the line to be tangent to the circle, this quadratic equation must have exactly one solution for $x$ , meaning its discriminant must equal zero: $[2 (3 c - 7)]^{2} - 4 (10) (c^{2} - 4 c - 4) = 0$

Expand the discriminant: $4 (9 c^{2} - 42 c + 49) - 40 (c^{2} - 4 c - 4) = 0$

Divide both sides by 4: $9 c^{2} - 42 c + 49 - 10 c^{2} + 40 c + 40 = 0$

Simplify: $- c^{2} - 2 c + 89 = 0$

Multiply by $- 1$ : $c^{2} + 2 c - 89 = 0$

Solve for $c$ using the quadratic formula: $c = \frac{- 2 \pm ( 2 ) ^{2} - 4 ( 1 ) ( - 89 )}{2 ( 1 )} = \frac{- 2 \pm 4 + 356}{2} = \frac{- 2 \pm 360}{2}$

Simplify $360 = 36 \times 10 = 610$ : $c = \frac{- 2 \pm 6 10}{2} = - 1 \pm 310$

Substituting these values of $c$ back into the line equation $y = 3 x + c$ :

$y = 3 x + (- 1 \pm 310)$

Or equivalently: $y = 3 x - 1 + 310 and y = 3 x - 1 - 310$

Method 2: Geometric Approach (Using Distance Formula)

The circle has center $(1, 2)$ and radius $r = 3$ . For a line to be tangent to a circle, the perpendicular distance from the center to the line must equal the radius.

Using the distance formula from point $(x_{0}, y_{0}) = (1, 2)$ to line $3 x - y + c = 0$ : $\frac{∣3 ( 1 ) - ( 2 ) + c ∣}{3 ^{2} + ( - 1 ) ^{2}} = 3$

Simplify: $\frac{∣1 + c ∣}{10} = 3$

Multiply both sides by $10$ : $∣1 + c ∣ = 310$

This gives two cases: $1 + c = 310 or 1 + c = - 310$

Solving for $c$ : $c = - 1 + 310 or c = - 1 - 310$

Substituting back into the standard form equation $3 x - y + c = 0$ :

$3 x - y + (- 1 \pm 310) = 0$

Or equivalently: $3 x - y - 1 + 310 = 0 and 3 x - y - 1 - 310 = 0$

Key Formulas or Methods Used

Perpendicular Slopes: If two lines with slopes $m_{1}$ and $m_{2}$ are perpendicular, then $m_{1} \cdot m_{2} = - 1$ (or $m_{2} = - \frac{1}{m _{1}}$ )
Algebraic Tangency Condition: A line is tangent to a circle if the system of equations yields a quadratic with discriminant $D = b^{2} - 4 a c = 0$
Distance from Point to Line: $d = \frac{∣ A x _{0} + B y _{0} + C ∣}{A ^{2} + B ^{2}}$ for line $A x + B y + C = 0$ and point $(x_{0}, y_{0})$
Geometric Tangency Condition: Distance from circle's center to the line equals the radius ( $d = r$ )

Summary of Steps

Find the tangent slope: Calculate the negative reciprocal of the given line's slope ( $- \frac{1}{3}$ ) to get $m = 3$
Write line equation: Express the tangent line as $y = 3 x + c$ (or $3 x - y + c = 0$ ) with unknown $c$
Apply tangency condition:
- Method 1: Substitute into circle equation, set discriminant of resulting quadratic to zero, and solve for $c$
- Method 2: Set distance from center $(1, 2)$ to line equal to radius $3$ , solve $∣1 + c ∣ = 310$ for $c$
Solve for $c$ : Obtain $c = - 1 \pm 310$
Write final equations: Substitute $c$ values back to get the two tangent lines: $y = 3 x - 1 \pm 310$ or $3 x - y - 1 \pm 310 = 0$

7.2 Q-3

Question Statement

Background and Explanation

Solution

Method 1: Algebraic Approach (Using the Discriminant)

Method 2: Geometric Approach (Using Distance Formula)

Key Formulas or Methods Used

Summary of Steps

7.2 Q-3

Question Statement

Background and Explanation

Solution

Method 1: Algebraic Approach (Using the Discriminant)

Method 2: Geometric Approach (Using Distance Formula)

Key Formulas or Methods Used

Summary of Steps