This is the question:

1. Point P has coordinates (2, k) and lies on the circumference of a circle with the equation

x^2 + y^2 − 3x + 2y + 2 = 0

(i) Verify that one possible value for k is 0 and find the other possible value for k. [2]

(ii) Find the equation of the tangent to the circle at P given that k is negative. [5]

(iii) Find where the tangent to the circle at P crosses the coordinate axes. [2]

1. Point P has coordinates (2, k) and lies on the circumference of a circle with the equation

x^2 + y^2 − 3x + 2y + 2 = 0

(i) Verify that one possible value for k is 0 and find the other possible value for k. [2]

(ii) Find the equation of the tangent to the circle at P given that k is negative. [5]

(iii) Find where the tangent to the circle at P crosses the coordinate axes. [2]

Original post by vc94

(i) Were you able to verify that k=0 works?

No. I'm not really sure where to start.

First start with completing the square to put it into the form (x+a)^2 + (y+b)^2 = r^2

Then, you can find the values of k for which the equation of the circle works(by finding the value of y that fits the value of x = 2.

To find the equation of a tangent, use the coordinate of ur negative value of k and the centre of the circle(given as (-a,-b)) to find change in y over change in x for gradient. then do the negative reciprocal of this for gradient of tangent(since it is perpendicular) and use y-y1 = m(x-x1) to get equation of tangetn.

u should be able to find out when it crosses coordinate axes pretty easily after that

Then, you can find the values of k for which the equation of the circle works(by finding the value of y that fits the value of x = 2.

To find the equation of a tangent, use the coordinate of ur negative value of k and the centre of the circle(given as (-a,-b)) to find change in y over change in x for gradient. then do the negative reciprocal of this for gradient of tangent(since it is perpendicular) and use y-y1 = m(x-x1) to get equation of tangetn.

u should be able to find out when it crosses coordinate axes pretty easily after that

Original post by Aga19

No. I'm not really sure where to start.

(i) When k=0, the point is (2, k). So just substitute x=2 and y=0 into the equation and see if it is satisfied... hence verified.

To find the other k value, just substitute x=2 and y=k into the equation, simplify to get a quadratic and solve for k.

Original post by Harry Bishop

First start with completing the square to put it into the form (x+a)^2 + (y+b)^2 = r^2

Then, you can find the values of k for which the equation of the circle works(by finding the value of y that fits the value of x = 2.

To find the equation of a tangent, use the coordinate of ur negative value of k and the centre of the circle(given as (-a,-b)) to find change in y over change in x for gradient. then do the negative reciprocal of this for gradient of tangent(since it is perpendicular) and use y-y1 = m(x-x1) to get equation of tangetn.

u should be able to find out when it crosses coordinate axes pretty easily after that

Then, you can find the values of k for which the equation of the circle works(by finding the value of y that fits the value of x = 2.

To find the equation of a tangent, use the coordinate of ur negative value of k and the centre of the circle(given as (-a,-b)) to find change in y over change in x for gradient. then do the negative reciprocal of this for gradient of tangent(since it is perpendicular) and use y-y1 = m(x-x1) to get equation of tangetn.

u should be able to find out when it crosses coordinate axes pretty easily after that

Thank you, this helped a lot.

Original post by vc94

(i) When k=0, the point is (2, k). So just substitute x=2 and y=0 into the equation and see if it is satisfied... hence verified.

To find the other k value, just substitute x=2 and y=k into the equation, simplify to get a quadratic and solve for k.

To find the other k value, just substitute x=2 and y=k into the equation, simplify to get a quadratic and solve for k.

Thank you.

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