# circles helpWatch

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#1
so what ive done is i expanded the circle equation to get x^2-6x+y^2+6y-35=0 and then i substituted y=3/2x+c into the circle to eliminate y and got x^2-6x+9x^2/3+3xc+c^2+9x+6c
i dont know what to do at this point because i have 2 unknowns and i cant really use the discriminant because its not a quadratic. anyone help ?
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1 month ago
#2
the intersections are going to be the intersections of the circle with a line of gradient -2/3 that passes through the centre, which is a little easier to manage, but what you have there is a quadratic, find the discriminant and you should end up with a quadratic in c, because there are two possible values for c that cause a gradient 3/2 line to just touch the circle
Last edited by Meowstic; 1 month ago
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#3
i dont get what you mean, becuase ive reattempted this question and used y=-2/3x+c and subbed it into x^2-6x+9+y^2+6y+9=52
and i got x^2-6x+9+4x^2/9-4xc/3+c^2-4x+6c+9=52
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1 month ago
#4
(Original post by 13.oswin.tsang)
i dont get what you mean, becuase ive reattempted this question and used y=-2/3x+c and subbed it into x^2-6x+9+y^2+6y+9=52
and i got x^2-6x+9+4x^2/9-4xc/3+c^2-4x+6c+9=52
This is very similar to the last question I helped you with. You need to group the x^2, x terms etc. like you did last time. Then use the discriminant.
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#5
(Original post by Notnek)
This is very similar to the last question I helped you with. You need to group the x^2, x terms etc. like you did last time. Then use the discriminant.
im stuck on this bit because theres a c^2 in the equation and i dont know how to group it. like i know that im meant to factorise to get it in the form ax^2+bx+c where a is what i factorised but i dont know how to do it with the c^2
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1 month ago
#6
(Original post by 13.oswin.tsang)
im stuck on this bit because theres a c^2 in the equation and i dont know how to group it. like i know that im meant to factorise to get it in the form ax^2+bx+c where a is what i factorised but i dont know how to do it with the c^2
c^2 is a constant like 2 so group it with all the other numbers. So your ‘c’ could be something like

(c^2 + 5)
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#7
(Original post by Notnek)
This is very similar to the last question I helped you with. You need to group the x^2, x terms etc. like you did last time. Then use the discriminant.
ive just searched it up using the solution bank and ive just done it completely wrong, thats why i got c as a unknown..i shouldve used y=-2/3x+c and subbed the circle midpoint to find the equation of the diameter of the circle that is perpendicular to l1 and l2 then to solve it
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#8
(Original post by Notnek)
c^2 is a constant like 2 so group it with all the other numbers. So your ‘c’ could be something like

(c^2 + 5)
my equation that i have is 13x^2/9+c^2-2x-4xc/3+6c+9+9-52
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#9
(Original post by Notnek)
c^2 is a constant like 2 so group it with all the other numbers. So your ‘c’ could be something like

(c^2 + 5)
could i ask, when would u need to know to use the discriminant
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1 month ago
#10
I gave the question a try;
You usually never need to expand the equation of a circle. Instead, since it tells you the gradient of the tangents, you can work out the gradient of the line that passes through the centre and the two tangents (perpendicular). Then you can use that equation to find the intersection and allowing you to complete the rest of the question.

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#11
(Original post by ShylaB14)
I gave the question a try;
You usually never need to expand the equation of a circle. Instead, since it tells you the gradient of the tangents, you can work out the gradient of the line that passes through the centre and the two tangents (perpendicular). Then you can use that equation to find the intersection and allowing you to complete the rest of the question.

yeah thats how i ended up doing it at the end, because the first method i thought of didnt work. thanks !
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1 month ago
#12

But you must be able to handle the algebra effectively, too.

There are many questions in the text book and online for you to see examples of where/how/why to use the discriminant in problems like this.
(Original post by 13.oswin.tsang)
yeah thats how i ended up doing it at the end, because the first method i thought of didnt work. thanks !
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