Question

The tangent lines to the circle x^2+y^2-6x+4y=12

Answer 1

Answer:

For the circle ... x² + y² - 6x + 4y - 12 = 0,

center C ≡ ( -6/-2, 4/-2 ) ≡ ( 3, -2),

radius r = √(9+4+12) = 5.

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A tangent parallel to line 4x + 3y + 5 = 0 has equation

4x + 3y + c = 0 .......... (1)

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If seg CP is perpendicular to this tgt, then

CP = r

| 4(3) + 3(-2) + c } / √( 4² + 3² ) = 5

| c + 6 | = 25

c + 6 = ± 25

= -6 ± 25

c = 19, -31........... (2)

From (1) and (2), the req'd tgt lines are

4x + 3y + 19 = 0 ... and ... 4x + 3y - 31 = 0 .

Answer 2

Answer:

Step-by-step explanation: