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Neep help with a quadratics question!! (1 Viewer)

JasonNg

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Can someone please show me how to do this question?

Consider the circle x^2 + y^2 -2x - 14y + 25 = 0,

(i) Show that if the line y=mx intersects the circle in two distinct points, then (1 + 7m)^2 - 25(1 + m^2) > 0

(ii) For what values of m is the line y=mx a tangent to the circle?

Thanks so much if you could show the working out :)
 

Carrotsticks

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Part i:

1. Sub y=mx into the equation of the circle.

2. Let the discriminant be greater than 0 (for 2 roots)

3. Acquire answer.

Part ii:

1. Use the discriminant from the previous part and let it be equal to 0.

2. Solve for m.
 

cutemouse

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In 2U students will probably find it easier to use the perpendicular distance formula.

That is,

Find the centre of the circle and the radius (by completing the square) and

(i) The perpendicular distance from the centre of the circle to the line y=mx should be less than the radius of the circle

(ii) The perpendicular distance from the centre of the circle to the line y=mx should be equal to the radius of the circle

Hope that helps.

Can someone please show me how to do this question?

Consider the circle x^2 + y^2 -2x - 14y + 25 = 0,

(i) Show that if the line y=mx intersects the circle in two distinct points, then (1 + 7m)^2 - 25(1 + m^2) > 0

(ii) For what values of m is the line y=mx a tangent to the circle?

Thanks so much if you could show the working out :)
 

JasonNg

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Can you guys please help me with one more question?

If one of the roots of the equation 2x^2 + 3kx = 5 is 4, find the values of k and the other root.

Thanks!
 

Carrotsticks

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Use the sum of roots.

Sum of roots = -3k/2 = A + B

But we know that one of the roots is 4.

-3k/2 = 4 + B (call this expression #1)

Now use the product of roots:

-5/2 = AB = 4B

Hence we know B = -5/8.

Use expression #1 to solve for K.

ALTERNATIVELY:

1. Substitute x=4 into the polynomial and let it equal 0, so P(4) = 0.

2. Solve for K.

3. Use sum of roots (taking note one of them is 4) to find the other root.
 

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