# circles,tangents problem...

• Dec 13th 2011, 06:32 AM
GalXiOn
circles,tangents problem...
hello guys a new member here..
i need help with this math problem that i got recently... its in the chapter conic sections.
the question is:
Find the equation of the tangent to the circle
if the equation of the circle is give by x^2+y^2-4x+6y-12=0 and is parallel to the line having the equation given by x+y-8=0

any help would be appreciated :)..
cheers.
EDIT: last question was wrong and i updated it with a valid question.
thx for the response.
• Dec 13th 2011, 07:43 AM
Plato
Re: circles,tangents problem...
Quote:

Originally Posted by GalXiOn
Find the equation of the tangent to the circle
Given: eq. of circle - x^2-4x-6y-12=0 and tangent is parallel to the line given by the equation 4x-3y=1.

There is something wrong with the question.
$x^2-4x-6y-12=0$ is not a circle.
Please check the wording of that question.
• Dec 13th 2011, 08:22 AM
GalXiOn
Re: circles,tangents problem...
Quote:

Originally Posted by Plato
There is something wrong with the question.
$x^2-4x-6y-12=0$ is not a circle.
Please check the wording of that question.

umm in my book it says if the equation if of the form x^2+y^2+2gx+2fy+c=0 then it is a General form i.e it represents the circle.
• Dec 13th 2011, 08:25 AM
Plato
Re: circles,tangents problem...
Quote:

Originally Posted by GalXiOn
umm in my book it says if the equation if of the form x^2+y^2+2gx+2fy+c=0 then it is a General form i.e it represents the circle.

But what you posted is not a circle there is no $y^2~.$
• Dec 13th 2011, 08:26 AM
GalXiOn
Re: circles,tangents problem...
Quote:

Originally Posted by Plato
But what you posted is not a circle there is no $y^2~.$

EDIT: updated.... (question given was wrong Sorry.. updated now...)
• Dec 13th 2011, 09:13 AM
GalXiOn
Re: circles,tangents problem...
x^2+y^2-4x+6y-12=0 and parallel to line having the equation x+y-8=0.. will this work?.. im just listing out a few sums that i got recently.. ive never seen these sorta sums before.. and are totally new.. im guessing its gotta do something with comparing the radius using 2 different formulas to get a variable.. something like equating (|ax1+by1+c|)/ sqrt(A^2+b^2) which is the distance from a line and a point
and this formula: sqrt(g^2+f^2-c) which is the radius of the circle if g,f,c is known to us(which can be found using the given equation) g is -2 and f is -3..(im not sure..)
now i dont know how to proceed or i dont know if it is even the right way to approach.. ..
• Dec 13th 2011, 09:31 AM
Plato
Re: circles,tangents problem...
Quote:

Originally Posted by GalXiOn
x^2+y^2-4x+6y-12=0 and parallel to line having the equation x+y-8=0..

By completing squares that circle is $(x-2)^2+(y+3)^2=25$.
The slope of any non-vertical tangent is $m=\frac{-(a-2)}{(b+3)}$ where $(a-2)^2+(b+3)^2=25$ and $b\ne -3$.

The slope of the line $x+y-8=0$ is $-1~.$

So find points $(a,b)$ where slope is $-1$.
• Dec 13th 2011, 11:29 PM
GalXiOn
Re: circles,tangents problem...
Quote:

Originally Posted by Plato
By completing squares that circle is $(x-2)^2+(y+3)^2=25$.
The slope of any non-vertical tangent is $m=\frac{-(a-2)}{(b+3)}$ where $(a-2)^2+(b+3)^2=25$ and $b\ne -3$.

The slope of the line $x+y-8=0$ is $-1~.$

So find points $(a,b)$ where slope is $-1$.

hmm thx i'll try it out..