Question

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).

Answer 1

well there is a formula for the perpendicular distance from a point to a line...

in this case the distance would be the radius...

Answer 2

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

Answer 3

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

Answer 4

thanks :)

Answer 5

But how did you get the denominator for d?

Answer 6

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

Answer 7

Okay thanks!

Answer 8

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).