NCERT Solutions: Exercise 10.1 - Conic Sections

(x­ - h)² + (y ­- k)² = r²
It is given that centre (h, k) =

and radius (r) = .Therefore, the equation of the circle is
Q4: Find the equation of the circle with centre (1, 1) and radius 

Ans: The equation of a circle with centre (h, k) and radius r is given as
(x­ - h)² + (y ­- k)² = r²
It is given that centre (h, k) = (1, 1) and radius (r) =Therefore, the equation of the circle is
Q5: Find the equation of the circle with centre (-a, -b) and radius 

Ans: The equation of a circle with centre (h, k) and radius r is given as
(x­ - h)² + (y ­- k)² = r²
It is given that centre (h, k) = (-a, -b) and radius (r) = .Therefore, the equation of the circle is 
Q6: Find the centre and radius of the circle (x + 5)² + (y - 3)² = 36
Ans: The equation of the given circle is (x + 5)² + (y - 3)² = 36.
Rewrite as (x - (-5))² + (y - 3)² = 6², which is of the form (x - h)² + (y - k)² = r².
Therefore h = -5, k = 3 and r = 6.
Thus, the centre is (-5, 3) and the radius is 6.
Q7: Find the centre and radius of the circle x² +  y² - 4x - 8y - 45 = 0
Ans: The equation of the given circle is x² +  y² - 4x - 8y - 45 = 0.
Group terms and complete the square:
(x² - 4x) + (y² - 8y) = 45
Complete squares:
{x² - 4x + 4} + {y² - 8y + 16} = 45 + 4 + 16
⇒ (x - 2)² + (y - 4)² = 65.
Thus h = 2, k = 4 and r² = 65, so r = √65.
Thus, the centre is (2, 4) and the radius is √65.
Q8: Find the centre and radius of the circle x² +  y² - 8x+  10y - 12 = 0
Ans: The equation of the given circle is x² +  y² - 8x + 10y - 12 = 0.
Group and complete the square:
(x² - 8x) + (y² + 10y) = 12
Add and subtract squares:
{x² - 8x + 16} + {y² + 10y + 25} = 12 + 16 + 25
⇒ (x - 4)² + (y + 5)² = 53.
Therefore h = 4, k = -5 and r= √53.Thus, the centre is (4, -5) and the radius is √53.
Q9: Find the centre and radius of the circle 2x² + 2y² - x = 0
Ans: The equation of the given circle is 2x² + 2y² - x = 0.
Divide both sides by 2:
x² + y² - (1/2)x = 0
Rewrite the x-terms and complete the square:
x² - (1/2)x + y² = 0
Add and subtract (1/4)² = 1/16 to complete square in x:
{x² - (1/2)x + 1/16} + y² = 1/16
⇒ (x - 1/4)² + y² = (1/4)².
Hence the centre is (1/4, 0) and the radius is 1/4.(h = 1/4, k = 0, r= 1/4).Thus, the centre isand the radius isQ10: Find the equation of the circle passing through the points (4, 1) and (6, 5) and whose centre is on the line 4x +  y = 16.
Ans: Let the required circle be (x - h)² + (y - k)² = r².
Since it passes through (4,1) and (6,5):
(4 - h)² + (1 - k)² = r² ... (1)
(6 - h)² + (5 - k)² = r² ... (2)
Centre (h, k) lies on 4x + y = 16:
4h + k = 16 ... (3)
Subtract (1) from (2):
(6 - h)² - (4 - h)² + (5 - k)² - (1 - k)² = 0
Compute differences:
(36 - 12h + h²) - (16 - 8h + h²) + (25 -10k + k²) - (1 -2k + k²) = 0
⇒ (20 - 4h) + (24 - 8k) = 0 ⇒ 4h + 8k = 44 ⇒ h + 2k = 11 ... (4)
Solve (3) and (4):
4h + k = 16
h + 2k = 11
Multiply second by 4 and subtract first×1 to eliminate h or solve directly:
From (4): h = 11 - 2k. Substitute in (3): 4(11 - 2k) + k = 16 ⇒ 44 - 8k + k = 16 ⇒ -7k = -28 ⇒ k = 4 ⇒ h = 3.
Substitute in (1): (4 - 3)² + (1 - 4)² = r² ⇒ 1 + 9 = r² ⇒ r² = 10.
Therefore the circle is (x - 3)² + (y - 4)²= 10.Expand if required:
x² + y² - 6x - 8y+ 15 = 0.
Q11: Find the equation of the circle passing through the points (2, 3) and (-1, 1) and whose centre is on the line x - 3y - 11 = 0.
Ans: Let the required circle be (x - h)² + (y - k)² = r².
Since it passes through (2,3) and (-1,1):
(2 - h)² + (3 - k)² = r² ... (1)
(-1 - h)² + (1 - k)² = r² ... (2)
Centre (h, k) lies on x - 3y - 11 = 0 ⇒ h - 3k = 11 ... (3)
Subtract (2) from (1):
(2 - h)² - (-1 - h)² + (3 - k)² - (1 - k)² = 0
Compute differences:
(4 - 4h) - (1 + 2h) + (9 - 6k) - (1 - 2k) = 0
⇒ (4 - 4h - 1 - 2h) + (9 - 6k - 1 + 2k) = 0
⇒ (3 - 6h) + (8 - 4k) = 0 ⇒ -6h - 4k +11 = 0 ⇒ 6h + 4k = 11 ... (4)
Solve (3) and (4):
h - 3k = 11
6h + 4k = 11
Multiply first by 6: 6h - 18k = 66. Subtract the second: (6h - 18k) - (6h + 4k) = 66 - 11 ⇒ -22k = 55 ⇒ k = -55/22 = -5/2.
Then h = 11 + 3k = 11 + 3(-5/2) = 11 - 15/2 = 22/2 - 15/2 = 7/2.
Thus centre (h, k) = (7/2, -5/2).
Find r² using (1):
(2 - 7/2)² + (3 + 5/2)² = r²
(-3/2)² + (11/2)² = 9/4 + 121/4 = 130/4 = 65/2.
Therefore the circle is (x - 7/2)² + (y + 5/2)²= 65/2.Equivalently one may multiply through by 2 to obtain an integer-coefficient form.
Q12: Find the equation of the circle with radius 5 whose centre lies on x-axis and passes through the point (2, 3).
Ans: Let the required circle be (x - h)² + (y - k)² = r².
Centre lies on the x-axis ⇒ k = 0 and r = 5.
So (x - h)² + y² = 25. The circle passes through (2,3):
(2 - h)² + 3² = 25 ⇒ (2 - h)² = 16 ⇒ 2 - h = ±4.
If 2 - h = 4 ⇒ h = -2, giving (x + 2)² + y² = 25 ⇒ x² + y² + 4x - 21 = 0.
If 2 - h = -4 ⇒ h = 6, giving (x - 6)² + y² = 25 ⇒ x² + y² - 12x + 11 = 0.
Hence there are two such circles.
Q13: Find the equation of the circle passing through (0, 0) and making intercepts a and b on the coordinate axes.
Ans: Let the required circle be (x - h)² + (y - k)² = r².
Since it passes through (0,0): h² + k² = r².
Therefore equation becomes (x - h)² + (y - k)² = h² + k².
The circle passes through (a, 0) and (0, b). From (a, 0):
(a - h)² + (-k)² = h² + k² ⇒ a² - 2ah = 0 ⇒ h = a/2 (since a≠ 0).From (0, b):
(-h)² + (b - k)² = h² + k² ⇒ b² - 2bk = 0 ⇒ k = b/2 (since b≠ 0).
Thus centre is (a/2, b/2) and r² = (a/2)² + (b/2)² = (a² + b²)/4.
Therefore the equation is (x - a/2)² + (y - b/2)² = (a² + b²)/4.
Q14: Find the equation of a circle with centre (2, 2) and passes through the point (4, 5).
Ans: The centre is (h, k) = (2, 2).
Radius squared = distance² between (2,2) and (4,5):
(4 - 2)² + (5 - 2)² = 2² + 3²= 4 + 9 = 13.Thus the equation is
(x - 2)² + (y - 2)²= 13.
Q15: Does the point (-2.5, 3.5) lie inside, outside or on the circle x² +  y² = 25?
Ans: The given circle is x² + y² = 25, so its centre is (0, 0) and radius = 5.
Distance of (-2.5, 3.5) from (0, 0) = √[(-2.5)² + 3.5²] = √[6.25 + 12.25] = √18.5 = √(37/2) ≈ 4.30.Since this distance (≈ 4.30) is less than the radius 5, the point (-2.5, 3.5) lies inside the circle.