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URGENT HELP NEEDED- Equations of a Circle

The straight line y = ax - 6
touches the circle with the equation (x-7)^2 + (y-3)^2 = 2 at only one point.

Find the 2 possible values of a

Help would be HUGELY APPRECIATED

Reply 1

vc94

Substitute the line into the circle equation, simplify to get a quadratic equation in x.
The line is a tangent to the circle, so what does that tell you about the discriminant of the equation?

Reply 2

Flxtcherr

I struggled reading that

(edited 3 years ago)

Reply 3

gstudent12345

Original post by vc94

Substitute the line into the circle equation, simplify to get a quadratic equation in x.
The line is a tangent to the circle, so what does that tell you about the discriminant of the equation?

I know discriminant must equal 0 if there’s only one point but I’m really not sure, could you show me what to do?

Reply 4

vc94

Original post by gstudent12345

I know discriminant must equal 0 if there’s only one point but I’m really not sure, could you show me what to do?

(x-7)^2 + (ax - 6 -3)^2 = 2
Simplify this to get a quadratic =0, make sure you get the coefficients right.
Use discriminant=0, to get 2 values for a.

Reply 5

gstudent12345

Original post by vc94

(x-7)^2 + (ax - 6 -3)^2 = 2
Simplify this to get a quadratic =0, make sure you get the coefficients right.
Use discriminant=0, to get 2 values for a.

I have quadratic as x^2(1+a) -2x(9a+7) + 128= 0 but unsure what values to use for discriminant

Reply 6

vc94

Original post by gstudent12345

I have quadratic as x^2(1+a) -2x(9a+7) + 128= 0 but unsure what values to use for discriminant

Assuming that's correct, then take A=1+a, B= -2(9a+7) and C=128
Solve B^2-4AC=0

Reply 7

gstudent12345

Original post by vc94

Assuming that's correct, then take A=1+a, B= -2(9a+7) and C=128
Solve B^2-4AC=0

Could you please solve it from the start, i am very stuck 🤯🤯

Reply 8

vc94

Your equation is not quite right .... x^2(1+a) -2x(9a+7) + 128= 0
Should be x^2(1+a^2) -2x(9a+7) + 128= 0
So A=1+a^2, B= -2(9a+7) and C=128
Solve B^2-4AC=0 works!

Reply 9

gstudent12345

Original post by vc94

Your equation is not quite right .... x^2(1+a) -2x(9a+7) + 128= 0
Should be x^2(1+a^2) -2x(9a+7) + 128= 0
So A=1+a^2, B= -2(9a+7) and C=128
Solve B^2-4AC=0 works!

is this the answer 1.728 and -0.1008

that’s what i’ve got i’m really stuck, is that right ?

Reply 10

gstudent12345

Original post by vc94

Your equation is not quite right .... x^2(1 a) -2x(9a 7) 128= 0
Should be x^2(1 a^2) -2x(9a 7) 128= 0
So A=1 a^2, B= -2(9a 7) and C=128
Solve B^2-4AC=0 works!

is the answer 1.728 and -0.1008?

sorry i didn’t mean to send that twice

(edited 3 years ago)

Reply 11

vc94