Ex 10.1
Last updated at April 17, 2024 by Teachoo
Ex 10.1, 15 (Introduction) Does the point (–2.5, 3.5) lie inside, outside or on the circle x2 + y2 = 25? We know that equation of circle is (x − h)2 + (y − k)2 = r2 where (h, k) is the centre & r is the radius of circle If for any point (a, b) (a − h) 2 + (b − k) 2 = r2 Then point (a, b) lies on the circle (a − h) 2 + (b − k) 2 < r2 Then point (a, b) lie inside the circle (iii) (a − h) 2 + (b − k) 2 > r2 Then point (a, b) lie outside the circle Ex 10.1, 15 (Method 1) Does the point (–2.5, 3.5) lie inside, outside or on the circle x2 + y2 = 25? Given equation x2 + y2 = 25 (x – 0)2 + (y – 0)2 = 52 We know that equation of circle is (x – h)2 + (y – k)2 = r2 Here, h = 0, k = 0 & r = 5 Hence Centre = (h, k) = (0, 0) Radius = r = 5 Finding Distance between centre (0, 0) & point (−2.5, 3.5) We know that distance between two point (x1, y1) & (x2, y2) is d = √((𝑥2−𝑥1)2+(𝑦2−𝑦1)2) Putting values = √((0−(−2.5))^2+(0 −3.5)2) = √((2.5)2+(−3.5)2) = √((2.5)2+(3.5)2) = √(6.25+12.25) = √18.50 < √25 < 5 Thus, distance between centre (0,0) & point (-2.5, 3.5) is √18.50 which is less than 5 (radius) Hence point (-2.5, −3.5) lies inside the circle Ex 10.1, 15 (Method 2) Does the point (–2.5, 3.5) lie inside, outside or on the circle x2 + y2 = 25? We know that equation of circle is (x − h)2 + (y − k)2 = r2 where (h, k) is the centre & r is the radius of circle If for any point (a, b) (a − h) 2 + (b − k) 2 = r2 Then point (a, b) lies on the circle (a − h) 2 + (b − k) 2 < r2 Then point (a, b) lie inside the circle (a − h) 2 + (b − k) 2 > r2 Then point (a, b) lie outside the circle Given that Equation of circle is x2 + y2 = 25 We need to find that Point (−2.5, 3.5) lie inside, outside or on the circle x2 + y2 = 25 Putting x = −2.5 & y = 3.5 in x2 + y2 = (−2.5)2 + (3.5)2 = 6.25 + 12.25 = 18.50 < 25 Thus, (−2.5)2 + (3.5)2 < 25 i.e. (a − h) 2 + (b − k) 2 < r2 ∴ Point (−2.5, 3.5) lie inside the circle x2 + y2 =25