Prove that the line ' ____' is a tangent to the circle ' ______=0' (1 Viewer)

[dux&src]

Prove that the line 3x+4y+21=0 is a tangent to the circle X^2 - 8x +y^2+4y-5 =0..

loll i tried solving simultaneously but i got stuck..and then i tried another method where i found the centre of the circle and the radius by completing the square after taking 5 to the RHS...


so can anyone help me out here..lol

Thanks, would appreciate any help.

 

[annabackwards]

Try making a quadratic formula using simultaneous equations and then show that the discriminant = 0

 

[Drongoski]

Prove that the line 3x+4y+21=0 is a tangent to the circle X^2 - 8x +y^2+4y-5 =0..

loll i tried solving simultaneously but i got stuck..and then i tried another method where i found the centre of the circle and the radius by completing the square after taking 5 to the RHS...


so can anyone help me out here..lol

Thanks, would appreciate any help.

You can subst for y = -3x/4 - 21/4 into the eqn of the circle, as anna suggested, and end up with a rather messy quadratic eqn in x; show that its discriminant = 0.

or


circle is: (x-4)² + (y+2)² = 5²

.: centre (4, -2) and radius 5

Perpendicular distance of centre from line = !3x4 +4x(-2) + 21!/sqrt(3^2 + 4^2) = 5

Therefore centre is 5 units away fron the line = its radius

.: line just touches the circle.

Hence line is tangent to circle.

 

[Timothy.Siu]

yeah, anna's way would probably be the way i would do it.

but, could u find the centre of the circle, find the perpendicular distance from the centre to the line, and if that distance is the radius, say that is it a tangent?

edit: thats what drogonski just posted lol.

 

[dux&src]

lolll thanks everyone for all your help..

@drongoski: yeap the second method was what i was doing and yeah i just forgot the perpendicular distance bit...
soo yeah sub the x and y values into the equation of the line and put it over sqrt 3^2+4^2

which comes out to be 25/5 =5

Thanks so much...

 

[dux&src]

lolll thanks everyone for all your help..

@drongoski: yeap the second method was what i was doing and yeah i just forgot the perpendicular distance bit...
soo yeah sub the x and y values into the equation of the line and put it over sqrt 3^2+4^2

which comes out to be 25/5 =5

Thanks so much...