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.

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.

$\displaystyle x^2-4x-6y-12=0$ **is not a circle**.

Please check the wording of that question.

Re: circles,tangents problem...

Quote:

Originally Posted by

**Plato** There is something wrong with the question.

$\displaystyle x^2-4x-6y-12=0$ **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.

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 $\displaystyle y^2~.$

Re: circles,tangents problem...

Quote:

Originally Posted by

**Plato** But what you posted is not a circle there is no $\displaystyle y^2~.$

EDIT: updated.... (question given was wrong Sorry.. updated now...)

Re: circles,tangents problem...

well if not that,what about this one? :

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.. ..

Re: circles,tangents problem...

Quote:

Originally Posted by

**GalXiOn** well if not that,what about this one? :

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 $\displaystyle (x-2)^2+(y+3)^2=25$.

The slope of any non-vertical tangent is $\displaystyle m=\frac{-(a-2)}{(b+3)}$ where $\displaystyle (a-2)^2+(b+3)^2=25$ and $\displaystyle b\ne -3$.

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

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

Re: circles,tangents problem...

Quote:

Originally Posted by

**Plato** By completing squares that circle is $\displaystyle (x-2)^2+(y+3)^2=25$.

The slope of any non-vertical tangent is $\displaystyle m=\frac{-(a-2)}{(b+3)}$ where $\displaystyle (a-2)^2+(b+3)^2=25$ and $\displaystyle b\ne -3$.

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

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

hmm thx i'll try it out..