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, ...

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..?

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 ...

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 ...

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

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)

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

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

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.

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 😭

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 😭

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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