Question

The center of a circle is (−2,3) and a point on the circumference is (−5,−1)
(i) Find the equation of the line joining the two points
(ii) Show that the radius of the circle is 5 units
(iii) Write down the equation of the circle
(iv) Determine the equation of the tangent to the circle at the point
(v) Find the points of intersection of the circle above with the circle
𝑥2+𝑦2+6𝑥−7𝑦−10=0 (6 marks)

Answer 1

(i) slope is 4/3, so y-3 = 4/3 (x+2)
(ii) r^2 = (-5+2)^2 + (-1-3)^2
(iii) (x+2)^2 + (y-3)^2 = r^2
(iv) the tangent line has slope -3/4, so use the point-slope form as in (i)
(v) just solve (iii) = x^2+y^2+6x-7y=10 (use elimination to get rid of x^2 and y^2)

Post your work if you get stuck

Answer 2

thanks, I got similar answers. I just didn't know how to answer (v)

Answer 3

ok, I was able to solve it. Could you help me with this one.

The points 𝐴(3,−1 ,2) ,𝐵(1 ,2 ,−4 ) and 𝐶 (−1 ,1 ,−2 ) are three vertices of a parallelogram ABCD.
(i) Show that the vector 𝑟 = −16𝑗 – 8𝑘 is perpendicular to the plan through 𝐴 ,𝐵 and 𝐶 .
(6 marks)
(ii) Find the cartesian equation of the plane through 𝐴 ,𝐵 and 𝐶 . (4 marks)

Answer 4

you might want to start here:

math.stackexchange.com/questions/2686606/equation-of-a-plane-passing-through-3-points

Answer 5

so for the very first question (v)
i got x = 1.41 and y=2.64
what did you get?