Question: Find the equation of the circle which passes through the point (5,8) and touches the straight lines x=1, x=9 and X-axis.

Solution:
In a circle,
Passing point (x,y) = (5,8)
It touches X-axis, and straight lines x = 1, and x = 9.

Plotting the given information in a graph, we get,


Diameter = (9-1) units = 8 units
Radius (r) = Diameter/2 = 4 units

When a circle touches X-axis, k = r. So, k = 4 units

We have,
(x,y) = (5,8); r = 4; k = 4
Using formula,
(x - h)² + (y - k)² = r²
or, (5 - h)² + (8 - 4)² = 4²
or, (5 - h)² + 16 = 16
or, (5 - h)² = 0
or, 5 - h = 0
So, h = 5 units
So, (h,k) = (5,4) and r = 4
Now, equation of a circle is given by,
(x - h)² + (y - k)² = r²
or, (x - 5)² + (y - 4)² = 4²
or, (x² - 10x + 25) + (y² - 8y + 16) + 16
or, x² + y² - 10x - 8y + 25 = 0 is the required equation.

Hence, the required equation of the given circle is x² + y² - 10x - 8y + 25 = 0.

#SciPiPupil