# im confused

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• Oct 8th 2009, 08:47 AM
Mathboy
im confused
helloo i need to get an answer for this question

Show that the line y = mx + c touches the circle x^2 + y^2 = a^2 if c^2 = a^2(1+m^2)

it's very hard for me (Thinking) i need some help plz.
• Oct 8th 2009, 10:14 AM
ramiee2010
Quote:

Show that the line y = mx + c touches the circle x^2 + y^2 = a^2 if c^2 = a^2(1+m^2)
\$\displaystyle x^2 + y^2 = a^2\$
\$\displaystyle x^2 + (mx + c)^2 = a^2\$
\$\displaystyle x^2 + m^2 x^2 + c^2+2mcx = a^2\$
\$\displaystyle (1 + m^2) x^2 +2mcx + (c^2- a^2)=0\$

if the line y = mx + c touches the circle(or tangent to the circle) then
\$\displaystyle discriminant = 0 \$
\$\displaystyle d= (2mc)^2 -4(1 + m^2)(c^2- a^2)=0\$
\$\displaystyle 4a^2-4c^2+4m^2a^2=0\$
\$\displaystyle c^2 = a^2+a^2m^2\$
\$\displaystyle c^2 = a^2(1+m^2)\$
• Oct 8th 2009, 10:24 AM
Mathboy
Any help about this also?

The point (p,q) lies on the circle x^2 + y^2 = a^2.
Show that the equation of the tangent to the circle at (p,q) is px + qy = a^2
• Oct 8th 2009, 10:55 AM
ramiee2010
the equation of the tangent to the circle having centre at origin is given by
\$\displaystyle x_{0}x+y_{0}y=r^2\$
\$\displaystyle where (x_{0},y_{0}) are\ point\ of\ intersection\$
\$\displaystyle since\ (p,q)\ lies\ on\ the\ circle\ x^2 + y^2 = a^2.\$
\$\displaystyle therefore\ p^2 + q^2 = a^2 =(radius)^2\$
therefore equation of the tangent
\$\displaystyle px+qy= p^2 + q^2 \$