## 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).

1. campbell_st

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

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

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

thanks :)

5. steve816

But how did you get the denominator for d?

6. campbell_st

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

Okay thanks!

8. IrishBoy123

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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