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steve816

  • one year ago

Please help me ASAP! How do I find the equation of the circle? The circle is tangent to the line y = 2x + 5 with center (5, -5).

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  1. campbell_st
    • one year ago
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    well there is a formula for the perpendicular distance from a point to a line... in this case the distance would be the radius...

  2. campbell_st
    • one year ago
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    so rewrite the line as 2x - y + 5 = 0 then the formula is \[d = \frac{|2x - y + 5|}{\sqrt{2^2 + (-1)^2}}\] just substitute the point (5, -5) to find the radius

  3. campbell_st
    • one year ago
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    then the equation will be \[(x - h)^2 + (y - k)^2 = r^2\] (h, k) is the centre, you have (5, -5) and r is the radius, in your case use d. hope it makes sense

  4. steve816
    • one year ago
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    thanks :)

  5. steve816
    • one year ago
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    But how did you get the denominator for d?

  6. campbell_st
    • one year ago
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    well is the equation of the tangent is in standard form Ax + By + C = 0 the formula is \[d = \frac{ \left| Ax + By + C \right|}{\sqrt{A^2 + B^2}}\] the denominator is from pythagoras' theorem. so all you do is just substitute you value of x = 5 and y = -5 and you get \[d = \frac{\left| 20 \right|}{\sqrt{5}}\] so that's the radius... square the numerator and denominator and you get \[d^2 = (\frac{20}{\sqrt{5}})^2~~~or~~~~d^2 = \frac{400}{5}\] then simplify that. I did it quickly so just check the calculations

  7. steve816
    • one year ago
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    Okay thanks!

  8. IrishBoy123
    • one year ago
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    you can also tell that the normal to the circle at the intersection is of the form \(y = -\frac{1}{2} x + c \) as it's slope will be -1/m for a tangent with slope m thusly, use the centre (5,-5) of the circle to solve for c, as the normal also runs through the centre of the circle so \(-5 = -\frac{1}{2} (5) + c \). then solve \(-\frac{1}{2} x + c\ = 2x + 5\) to find the point where the circle actually intersects the tangent line. the circle's radius will be the distance from that point to the centre of circle at (5,-5). |dw:1443694428546:dw|

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