Question

Find the equation of the tangent
to the circle x² + y ² - 4x+6y - 12 = 0
which are parallel to x+y-8-0​

Answer 1

Answer:

Equation of the tangent which is parallel to

x+ y -8 =0 will be in the form of x+y+k = 0

centre of the circle = ( 2, -3)

radius =

$\sqrt{ {2}^{2} + { (- 3)}^{2} -( - 12) }$

$\sqrt{4 + 9 + 12}$

$\sqrt{25}$

radius = 5

distance from ( 2, -3) to the line x+y+k = 0 is equal to radius of the circle

$\frac{ |2 - 3 + k| }{ \sqrt{13} } = 5$

$|k - 1| = 5 \sqrt{13}$

$k - 1 = + or - 5 \sqrt{13}$

$k \: = 5 \sqrt{13 } + 1 \: or \: - 5 \sqrt{13} + 1$

equation of the tangent is

$x + y + 5 \sqrt{13} + 1$

or

$x + y + - 5 \sqrt{13} + 1$

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