Question

Consider the circle x^2+y^2-6x+4y-12=0.The equation of a tangent to this circle that is parallel to the line 4x+3y+5=0.

Answer 1

please see the attachment

Answer 2

Answer:

$4x + 3y + 19 = 0 \\\\4x + 3y -31 = 0$

Step-by-step explanation:

We know that from the standard equation of the circle,one can find centre and radius of the circle

${x}^{2} + {y}^{2} + 2gx + 2fy + c = 0$

center (-g,-f) and radius

$r = \sqrt{ {g}^{2} + {f}^{2} - c}$

So,in the given equation

${x}^{2} + {y}^{2} - 6x + 4y - 12 = 0 \\ \\ center(3 ,- 2) \\ \\ r = \sqrt{9 + 4 + 12} \\ \\ = \sqrt{25} \\ \\ r = 5$

We also know that slope of parallel lines are equal and tangent touches the circle and make 90° angle with radius of the circle.

Hence equation of tangent would be

$4x + 3y + k = 0$

and perpendicular distance from center will equal to the radius of circle,so

apply the distance formula between a line and a point

$\bigg| \frac{ax + by + c}{ \sqrt{ { {x}^{2} } + {y}^{2} } } \bigg| \\ \\ \bigg| \frac{4(3) +( - 2)(3) + k}{ \sqrt{ { {4}^{2} } + {3}^{2} } } \bigg| = 5 \\ \\ \bigg| \frac{12 - 6 + k }{ \sqrt{25}} \bigg| = 5\\ \\ 6 + k =± 25 \\ \\ k = 25 - 6 \\ \\ k = 19 \\ \\or \\\\k=-25-6=-31$

Hence equation of tangent is

$4x + 3y + 19 = 0 \\\\4x + 3y -31 = 0$

Hope it helps you.