Q5. Find the circle with (−3, 2) and (5, −4) as the ends of a diameter.
(x + 3)(x − 5) + (y − 2)(y + 4) = 0, that is x2 + y2 − 2x + 2y − 23 = 0: centre (1, −1) and radius 5. Check: AB = √(64 + 36) = 10, twice the radius. This is the circle of Example 1; (−3, 2) and (5, −4) are the ends of another diameter.
Q6. Show that (5, 2) lies on the circle with (1, 2) and (5, 6) as the ends of a diameter, and find the angle these ends subtend at (5, 2).
(5 − 1)(5 − 5) + (2 − 2)(2 − 6) = 0, so (5, 2) is on the circle. From (5, 2), the point (1, 2) is straight to the left and (5, 6) straight up, so the angle is 90°, as the angle in a semicircle must be.
Q7. Find the length of the chord that y = 2x cuts off on x2 + y2 = 20.
The line passes through the centre (d = 0), so the chord is a diameter: 2√20 = 4√5. The ends are (2, 4) and (−2, −4).
Q8. For what values of c does y = 3x + c touch x2 + y2 = 10?
c2 = r2(1 + m2) = 10 × 10 = 100, so c = 10 or −10.
Q9. Find where 3x + 4y = 15 meets x2 + y2 = 25, and the length of the chord.
With y = (15 − 3x)/4: 25x2 − 90x − 175 = 0, that is 5x2 − 18x − 35 = 0, so x = 5 or −7/5. The points are (5, 0) and (−7/5, 24/5), and the chord is 2√(25 − 9) = 8 long, since d = 15/5 = 3.
Q10. Find k so that the line x + y = k touches the circle x2 + y2 = 18.
The line is y = −x + k, so k2 = 18 × 2 = 36 and k = 6 or −6. Check: the distance of the centre from x + y = 6 is 6/√2 = 3√2 = √18.