A circle has centre(5,12) and its tangent to the line with equation 2x-y+3=0. Wtite equation of the cirle
circle has centre at (5, 12) and is tangent to the line with equation 2x - y + 3 = 0 . Write the equation of the circle.
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Christian
That equation
Well, this circle seems to have a very tangential relationship with the line 2x-y+3=0. They're so close, they might as well be touching!
To find the equation of the circle, we need to determine its radius and its center. The center of the circle is given as (5,12), which means it's a very centered and balanced circle.
Now, let's focus on that line. Imagine trying to hold a circle and a line together - it'd be quite a delicate balancing act!
The equation of the line is 2x-y+3=0. If you rearrange it a bit, you'll find that y = 2x + 3. So, the slope of the line is 2. This line is definitely on a roll!
Since the circle is tangent to the line, that means it only touches the line at one point. It's like a game of tag - the circle is "it", and the line runs away as fast as it can, but the circle always manages to catch up and touch it at just one spot!
To find the radius, we need to find the distance between the center of the circle and the point of tangency. Luck for us, we know their coordinates! We'll call the point of tangency (a,b).
Now, let's use some geometry wizardry. The distance formula is our magic spell:
radius = sqrt((a-5)^2 + (b-12)^2)
Since the circle is only touching the line at one point, the radius is just enough to reach from the center to that point. It's like a friendly extension from the center of the circle, reaching out to say hello to the line.
Now, we have everything we need to write the equation of the circle:
(x-5)^2 + (y-12)^2 = radius^2
Just plug in the coordinates of the point of tangency, and voila! You've got yourself the equation of the circle. It's like solving a puzzle - everything fits together perfectly!
I hope this explanation wasn't too "circus-y" for you. If you need any more assistance, feel free to ask!
To find the equation of the circle, we need to know its radius. Since the circle is tangent to the given line, the distance from the center of the circle to the line will be equal to the radius of the circle.
Step 1: Convert the given equation of the line into slope-intercept form (y = mx + b).
2x - y + 3 = 0
-y = -2x - 3
y = 2x + 3
Step 2: Find the slope of the line.
The slope (m) of the line is the coefficient of x, which in this case is 2.
Step 3: Find the perpendicular slope to the line.
The slope of any line perpendicular to a line with slope m can be found by taking the negative reciprocal of m. So, the perpendicular slope (m_perpendicular) will be -1/m.
m_perpendicular = -1/2
Step 4: Use the perpendicular slope and the center of the circle to find the equation of the line passing through the circle's center and perpendicular to the given line.
Using the point-slope form of a line (y - y1 = m(x - x1)), where (x1, y1) is the center of the circle (5, 12), we can write:
y - 12 = -1/2(x - 5)
y - 12 = -1/2x + 5/2
y = -1/2x + 5/2 + 12
y = -1/2x + 29/2
Step 5: Find the intersection point of the two lines.
To find the intersection point of the two lines, we need to solve their equations simultaneously. Let's equate the two equations:
2x + 3 = -1/2x + 29/2
4x + 6 = -x + 29
5x = 23
x = 23/5
Substituting the value of x back into either of the line equations to find y:
y = -1/2(23/5) + 29/2
y = -23/10 + 145/10
y = 122/10
y = 61/5
So, the intersection point is (23/5, 61/5), which is the center of the circle.
Step 6: Find the distance between the center of the circle and the intersection point, which will give us the radius.
Using the distance formula:
radius = sqrt((x2 - x1)^2 + (y2 - y1)^2)
radius = sqrt((23/5 - 5)^2 + (61/5 - 12)^2)
radius = sqrt((23/5 - 25/5)^2 + (61/5 - 60/5)^2)
radius = sqrt((-2/5)^2 + (1/5)^2)
radius = sqrt(4/25 + 1/25)
radius = sqrt(5/25)
radius = sqrt(1/5)
radius = 1/sqrt(5)
Note: We simplify the square root by rationalizing the denominator.
Step 7: Write the equation of the circle using the center and radius.
(x - h)^2 + (y - k)^2 = r^2
(x - 23/5)^2 + (y - 61/5)^2 = (1/sqrt(5))^2
Multiplying both sides of the equation by 5, to get rid of the denominator:
5(x - 23/5)^2 + 5(y - 61/5)^2 = 1
Expanding the equation:
5(x^2 - (2 * 23x)/5 + (23/5)^2) + 5(y^2 - (2 * 61y)/5 + (61/5)^2) = 1
Simplifying:
5x^2 - 2 * 23x + 23^2 + 5y^2 - 2 * 61y + 61^2 = 1
5x^2 - 46x + 529 + 5y^2 - 122y + 3721 = 1
5x^2 - 46x + 5y^2 - 122y + 4249 = 1
Rearranging the terms:
5x^2 + 5y^2 - 46x - 122y + 4248 = 0
Thus, the equation of the circle is 5x^2 + 5y^2 - 46x - 122y + 4248 = 0.