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The line 10x + py = q is a tangent at the point (5, 4) in another circle with centre

(0,0).
Find the value of p and the value of q.

1 Answer

9 votes

Answer:

(p, q) = (8, 82)

Explanation:

When a circle is centered at the origin, the radius to point (a, b) will have slope m = b/a. The tangent is perpendicular to the radius, so the tangent at point (a, b) will have slope -a/b. In point-slope form, the equation of the tangent line will be ...

y -k = m(x -h) . . . . . point-slope equation of line with slope m through (h, k)

y -b = (-a/b)(x -a)

Rearranging this to standard form, we have ...

b(y -b) = -a(x -a)

by -b² = -ax +a²

ax +by = a² +b²

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For (a, b) = (5, 4), the standard form equation of the tangent can be written ...

5x +4y = 5² +4² = 41

Your given equation has an x-coefficient that is twice the value shown in this equation, so we need to multiply this equation by 2:

2(5x +4y) = 2(41)

10x +8y = 82

Comparing to 10x +py = q, we see that ...

p = 8

q = 82

The line 10x + py = q is a tangent at the point (5, 4) in another circle with centre-example-1
User Vyrotek
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