The equation of a given circle is x^2 + y^2 - 10x +16 = 0?

Question

Determine the values of k for which the line y=kx

i). cuts the given circle

ii). touches the given circle

iii). does not intersect the given circle

Pls help.. stuck..

Answer 1

The three parts of the answer start the same:

-replace y by it's value into the equation of the circle:

x2+(kx)2-10x+16=0

(k2+1)x2-10x+16=0

-find the discremenant of the second degree equation:

100-4*16*(k2-1)=

-64k2+164

i)cuts the circle=>dicremenant>0

k2<164/64

ii)touch the circle=>discemenant=0

k2=164/64

iii)does not intersect the circle=>discremenant<0

k2>164/64

Answer 2

x^2 +(kx)^2 -10x +16=0

(1+k^2)x^2 -10x +16=0

b^2 -4ac = (-10)^2 -4(1+k^2)(16)

100 -64(1+k^2) =

100 -64 -64k^2 =

36 -64k^2 >0 cuts the circle, -3/4 <k<3/4

36-64k^2=0 touches the circle, k=3/4 or -3/4

36-64k^2<0 does not intersect, k>3/4 or k<-3/4

64k^2<36

k^2 <36/64

Answer 3

The intersections, if there exist, are points whose coordinates are solutions to

x^2 + y^2 - 10x + 16 = 0 and

y = kx

or

x^2 + k^2.x^2 - 10x + 16 = 0

(1+k^2)x^2 - 10x + 16 = 0

Δ = 100 - 64(1 + k^2) = 36 - 64.k^2 = - (8k +6)(8k - 6)

..i) the line cuts the circle if Δ > 0, for -3/4 < k < +3/4

.ii) the line touches the circle if Δ = 0, for k = -3/4 or +3/4

iii) the line does not intersect the circle if Δ < 0, for k < -3/4 or k > +3/4

Answer 4

Eliminate y and form a quadratic in x.

If the D > 0 then (i) will happen

           = 0 then (ii) and if

           <0 then (iii)

Now give it a try!