Q5. Find the equation of the circle with centre (1/2, 1/4) and radius 1/12 (NCERT Exercise 10.1, Q3).
(x − 1/2)2 + (y − 1/4)2 = 1/144. Expanding, multiplying by 144 and dividing by 4:
36(x2 + y2) − 36x − 18y + 11 = 0.
Q6. Find the equation of the circle with centre (1, 1) and radius √2 (NCERT Exercise 10.1, Q4).
(x − 1)2 + (y − 1)2 = 2, that is x2 + y2 − 2x − 2y = 0. It passes through the origin.
Q7. Find the equation of the circle with centre (−a, −b) and radius √(a2 − b2) (NCERT Exercise 10.1, Q5).
(x + a)2 + (y + b)2 = a2 − b2, that is x2 + y2 + 2ax + 2by + 2b2 = 0 (for |a| > |b|).
Q8. Find the circle through (2, 3) and (−1, 1) whose centre is on x − 3y − 11 = 0 (NCERT Exercise 10.1, Q11).
The perpendicular bisector of the chord passes through (1/2, 2) with slope −3/2: 6x + 4y = 11. With x − 3y = 11 it gives the centre (7/2, −5/2), and r2 = (2 − 7/2)2 + (3 + 5/2)2 = 65/2.
The circle is x2 + y2 − 7x + 5y − 14 = 0.
Q9. Find the equation of the circle with centre (2, 2) passing through (4, 5) (NCERT Exercise 10.1, Q14).
r2 = 22 + 32 = 13, so (x − 2)2 + (y − 2)2 = 13, that is x2 + y2 − 4x − 4y − 5 = 0.
Q10. For what value of k does x2 + y2 + 4x − 6y + k = 0 have radius 5?
g = 2, f = −3, so the radius squared is 4 + 9 − k = 25 and k = −12.
Q11. Show that x2 + y2 − 2x + 4y + 5 = 0 is satisfied by only one point.
(x − 1)2 + (y + 2)2 = 1 + 4 − 5 = 0. A sum of two squares is zero only when both are zero, so the only point is (1, −2): a point circle.
Q12. Find the circle with centre (3, −4) that passes through the origin.
r2 = 9 + 16 = 25, so (x − 3)2 + (y + 4)2 = 25, that is x2 + y2 − 6x + 8y = 0. The constant term is 0 exactly because the circle passes through the origin.