Question

QUESTION: Find the constant k so that the line with equation y = kx is tangent to the circle with equation (x - 3)2 + (y - 5)2 = 4.
ANSWER:(x - 3)2 + (y - 5)2 = 4 : given
(x - 3)2 + (kx - 5)2 = 4 : substitute y by kx
x2(1 + k2) - x(6 + 10k) + 21 = 0 : expand and write quadratic equation in standard form.
(6 + 10k)2 - 4(1 + k2)(21) = 0 : For the circle and the line y = kx to be tangent, the discriminant of the above quadratic equation must be equal to zero.
16k2 + 120k - 48 = 0 : expand above equation
k = (-15 + sqrt(273)/4 , k = (-15 - sqrt(273)/4 : solve the above quadratic equation.

Answer 1

So are you looking to check your work?

Answer 2

Yes

Answer 3

@DamageGoods
\(9+25 -4 = 30 \)

Answer 4

Line 3
+10k should be -10k and +21 should be -23

Answer 5

+10k is correct actually

Answer 6

and how -23 ? @alekos ?

Answer 7

should be +30 instead of +21

Answer 8

Sorry +21 should be +30
Hartnn is right

Answer 9

Yes you're right.
Trying to do this in my head because I can't find a pen !

Answer 10

ok is that the only mistake?

Answer 11

you continue from there,
might happen some other mistake in later steps ...

Answer 12

so tell us what final values of k u get?
and we'll verify :)