The line 10x + py = q is a tangent at the point (5, 4) in another circle with centre
1 answer:
Answer:
(p, q) = (8, 82)
Step-by-step 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
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