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A level OCR maths question help

i dont even know where to start with this question
(edited 3 months ago)
Reply 1
Original post by esha06
i dont even know where to start with this question

A sketch should help. Youre given the radius is r and the question wants you to work the question through with r being a variable. So if it touches both axes, what would be the centre, ...
Reply 2
Original post by mqb2766
A sketch should help. Youre given the radius is r and the question wants you to work the question through with r being a variable. So if it touches both axes, what would be the centre, ...

would the equation be (x+r)^2+(x+y)^2=r^2 and then make that equal to 0 and make 2x+y=12 equal to 0 and then equate them..?
Reply 3
Original post by esha06
would the equation be (x+r)^2+(x+y)^2=r^2 and then make that equal to 0 and make 2x+y=12 equal to 0 and then equate them..?

Right idea for the circle, but check the centre.

You solve the line and circle simultaneously which as the circle is nonlinear/quadratic its easier to do by substitution. As you want to end up with a quadratic in x, you substitute ...
Reply 4
Original post by mqb2766
Right idea for the circle, but check the centre.

You solve the line and circle simultaneously which as the circle is nonlinear/quadratic its easier to do by substitution. As you want to end up with a quadratic in x, you substitute ...

whoops i meant (x+r)^2+(y+r)^2=r^2
Reply 5
Original post by esha06
whoops i meant (x+r)^2+(y+r)^2=r^2

Guessed that, but previous comment still applies. Your centre is not in Q1.
(edited 3 months ago)
Reply 6
Original post by mqb2766
Guessed that, but previous comment still applies.

idk whats wrong with itttt

when solving for x and y it would give you -r and doesn't that make sense since it has to be 1st quadrant
Reply 7
Original post by esha06
idk whats wrong with itttt

when solving for x and y it would give you -r and doesn't that make sense since it has to be 1st quadrant

x=y=-r would be in quadrant 3, so both negative. Check your notes for a circle.
Reply 8
Original post by mqb2766
x=y=-r would be in quadrant 3, so both negative. Check your notes for a circle.

ohhhh so it would be (x-r)^2+(y-r)^2=r^2
i was thinking about quadrant 2 and still did it wrong 😭
Reply 9
Original post by esha06
ohhhh so it would be (x-r)^2+(y-r)^2=r^2
i was thinking about quadrant 2 and still did it wrong 😭

Yup. If you know the centre so (r,r) here, you just check your circle equation gives
0^2 + 0^2 = ...
as the centre must be a zero distance from the centre. Hence you negate the actual sign of the centre in circle equation.

So now follow the simultaneous - substitution advice in a prevoius post.

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