Find the equation of circle touching the line 2x+3y+1=0
Answers
Answer:
One such circle has equation
x² + y² = 1 / 13
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Step-by-step explanation:
Since any circle that touches the line will do, make things simple and find the one with centre at the origin. Just need to work out the radius. This will be the (perpendicular) distance from the origin to the line.
The line, the x-axis and the y-axis together make a right angled triangle ABO (see the picture). Thinking of AB as the base, we need to work out the height OD, as this will be the radius of the circle that touches the line at D.
Find x- and y-intercepts of the line 2x + 3y + 1 = 0.
x-intercept: y = 0 => 2x + 1 = 0 => x = - 1 / 2
y-intercept: x = 0 => 3x + 1 = 0 => y = - 1 / 3
In ΔABO, we have AO = 1/2 and BO = 1/3.
By Pythagoras' Theorem,
AB = √(AO² + BO²) = √(1/4 + 1/9) = √(13/36) = √13 / 6
ΔABO and ΔOBD are similar
=> AO / AB = OD / OB
=> OD = AO × OB / AB
= (1/2) × (1/3) / (√13 / 6)
= 1 / √13
The equation of a circle is
( x - a )² + ( y - b )² = c²
where (a, b) is the centre and c is the radius.
Our circle has centre at (a, b) = (0, 0) and radius c = 1 / √13, so its equation is
x² + y² = 1 / 13
