A computer graphics program displays images on a rectangular screen whose coordinate system Class 9

A computer graphics program displays images on a rectangular screen whose coordinate system Class 9

Question 1.

A computer graphics program displays images on a rectangular screen whose coordinate system has the origin at the bottom-left corner. The screen is 800 pixels wide and 600 pixels high. A circular icon of radius 80 pixels is drawn with its centre at the point A (100, 150). Another circular icon of radius 100 pixels is drawn with its centre at the point B (250, 230). Determine:

i. whether any part of either circle lies outside the screen.

ii. whether the two circles intersect each other.

Solution:

We know that, if the end points of the diameter of the circle lie in the screen, then the entire circle will lie in the screen.

A = (100, 150) and radius = 80 pixels

Let the end points of the diameter be D, E, F and G.

∴ D = (100 + 80, 150) = (180, 150)

E = (100 - 80, 150) = (20, 150)

F = (100, 150 - 80) = (100, 70)

G = (100, 150 + 80) = (100, 230)

∴ The x and y coordinates are less than 800 and 600 respectively but not negative.

∴ The circle with centre A lies entirely in the screen.

B = (250, 230) and radius = 100 pixels

Let the end points of the diameter be P, Q, R and S.

∴ P = (250 + 100, 230) = (350, 230)

Q = (250 - 100, 230) = (150, 230)

R = (250, 230 - 100) = (250,130)

S = (250, 230 + 100) = (250, 330)

∴ The x and y coordinates are less than 800 and 600 respectively but not negative.

∴ The circle with centre B lies entirely in the screen

ii. Yes. The two circles intersect each other.

Distance between centres

$\begin{aligned} & =\sqrt{(250-100)^2+(230-150)^2} \\ & =\sqrt{(150)^2+(80)^2} \\ & =\sqrt{28900}\end{aligned}$

= 170

Sum of radii = 80 + 100 = 180

Since, 170 < 180

∴ The two circles intersect at two points.