Question

Find the equations of the tangents to the circlex^2 + y^2 - 4x + 6y - 12 = 0 which are parallel to
3x-y+4=0​

Answer 1

EXPLANATION.

→ Equation if tangent to the circle

x² + y² - 4x + 6y - 12 = 0.

→ parallel to the line = 3x - y + 4 = 0

→ x² + y² + 2gx + 2fy + c = 0 → general equation

of the circle.

→ compare the equation with general equation.

→ centre = ( -g, -f)

→ centre = ( 2,-3)

→ radius = √g² + f² - c

→ √ (2)² + (-3)² - (-12)

→ √ 4 + 9 + 12

→ √ 25 = 5

→ Tangent parallel to the line → 3x - y + 4 = 0

→ 3x - y + c = 0

$\sf : \implies \: | \dfrac{ax + by + c}{ \sqrt{ {a}^{2} + {b}^{2} } } | \\ \\ \sf : \implies \: | \frac{3(2) - ( - 3) + c}{ \sqrt{ {3}^{2} + {1}^{2} } } | = 5 \\ \\ \sf : \implies \: | \frac{6 + 3 + c}{ \sqrt{10} } | = 5 \\ \\ \sf : \implies \: 9 + c \: = 5 \sqrt{10}$

$\sf : \implies \: squaring \: on \: \: both \: sides \: \\ \\ \sf : \implies \: {9}^{2} + {c}^{2} = (5 \sqrt{10} ) {}^{2} \\ \\ \sf : \implies \: 81 + {c}^{2} = 250 \\ \\ \sf : \implies \: {c}^{2} = 250 - 81 \\ \\ \sf : \implies \: {c}^{2} = 169 \\ \\ \sf : \implies \: c \: = \pm \: 13$

→ equation of tangent.

→ 3x - y + 13= 0 and 3x - y - 13 = 0.

Answer 2

Equation if tangent to the circle

x² + y² - 4x + 6y - 12 = 0.

→ parallel to the line = 3x - y + 4 = 0

→ x² + y² + 2gx + 2fy + c = 0 → general equation

of the circle.

→ compare the equation with general equation.

→ centre = ( -g, -f)

→ centre = ( 2,-3)

→ radius = √g² + f² - c

→ √ (2)² + (-3)² - (-12)

→ √ 4 + 9 + 12

→ √ 25 = 5

→ Tangent parallel to the line → 3x - y + 4 = 0

→ 3x - y + c = 0





→ equation of tangent.
→ 3x - y + 13= 0 and 3x - y - 13 = 0