I don't know how to solve these two questions can someone help me out here?
Thanks in advance!
1. It is given that P(0, 3), Q(7, 5), R(9, -2) and S are vertices of a square. A circle A passes through points P, Q, R and S. Find the equation of the circle.
Hence or otherwise find the equation of circle B which is the image of A under the reflection about y-axis.
2. The straight line y = 5 - 2x meets the cirlce, C, x^2 + y^2 + 4x + - 8y - 5 = 0 at the points A and B. Find
(i) the coordinates of the centre, D, and the radius of the circle C,
(ii) the area of the triangle ADB.
1) eqn of ciecleA = (x-4.5)^2 + (y-0.5)^2 = 106
Eqn of circle B= (x+4.5)^2 + (y+0.5)^2 = 106
2) i) (-2,7); 5
ii) 4 unit^2
Personal answer....
not definate correct
I don't know but mine answer is similar to yours just that my radius is 26.5 instead of 106.
And how do i do it, like the steps in solving this question.
Hi,
In Q1, the centre of circle is the midpoint of P and R and the radius is half of the length of PR. Your value of 26.5 (square of the radius) is correct.
Thanks!
Cheers,
Wen Shih
Sorry... error in calculation~
ish indeed 26.5
Hi,
For Q2:
1. For a circle whose equation is given by x^2 + y^2 + 2gx + 2fy + c = 0, the centre is (-g, -f) and radius is sqrt(g^2 + f^2 - c).
2. Find points A and B, by substituting y = 5 - 2x into the equation of circle and solving a quadratic equation involving x.
3. Use the area formula (the one that requires cross-multiplications) to find triangle ADB, but be sure of their positions first (because vertices should be arranged anticlockwise or clockwise).
Thanks!
Cheers,
Wen Shih