[ASK] Equation of a Circle

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In summary, to find standard equations of circles with centers on 4x+3y=8 and tangent to both x+y=-2 and 7x-y=-6, we can use the equation (x-a)^2 + (y-(8-4a)/3)^2 = r^2 and set it equal to the line equations to find possible values for a and b. Then, using the quadratic formula, we can determine the conditions for the circles to be tangent to both lines.
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Monoxdifly
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find standard eqations of circles that have centers on 4x+3y=8 and are tangent to both the line x+y=-2 and 7x-y=-6

What I got is \(\displaystyle 4a=–4\pm3r\sqrt2\) and \(\displaystyle b=4\pm r\sqrt2\). Dunno how to continue from here.
 
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  • #2
Monoxdifly said:
find standard eqations of circles that have centers on 4x+3y=8 and are tangent to both the line x+y=-2 and 7x-y=-6

What I got is \(\displaystyle 4a=–4\pm3r\sqrt2\) and \(\displaystyle b=4\pm r\sqrt2\). Dunno how to continue from here.
I'm afraid this makes no sense because there are no "a" or "b" in the original question so we have no idea how this relates to the question.

Here's what I would do:
Let (a, b) be the center of such a circle. (I think that's what you intended but did not say.) Then, since the center lies on the line 4x+ 3y= 8, we have b= (8- 4a)/3. The circle can be written [tex](x- a)^2+ (y- (8-4a)/3)^2= r^2[/tex] and we need to find values for a and r. We have two more conditions.

If that circle crossed the line x+ y= 2 then the equation [tex](x- a)^2+ (2- x- (8- 4a/3)^2= r^2[/tex] would have 2 solutions. But if the line is tangent to the circle, that quadratic equation must have a double root. Do the indicated squares in
[tex](x- a)^2+ (2- x- (8- 4a/3)^2= r^2[/tex] to get the quadratic equation in standard form and use the quadratic formula to see what must be true about a and b.

Do the same with the line 7x- 4y= -6.
 
  • #3
Wow, didn't think that the steps were far simpler than I thought, though I still needed half a day to arrive at the final answer due to my lack of accuracy during calculating some things. Thanks for your help.
 
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I freaked out yesterday when my calculation got two centers of circle. Now I'm relieved that it does have two circles.
 

Related to [ASK] Equation of a Circle

What is the equation of a circle?

The equation of a circle is (x - h)^2 + (y - k)^2 = r^2, where (h,k) represents the center of the circle and r represents the radius.

How do you find the equation of a circle?

To find the equation of a circle, you need to know the coordinates of the center and the length of the radius. Then, you can plug these values into the equation (x - h)^2 + (y - k)^2 = r^2 to get the full equation.

What is the standard form of the equation of a circle?

The standard form of the equation of a circle is (x - h)^2 + (y - k)^2 = r^2, where (h,k) represents the center of the circle and r represents the radius. This form is also known as the center-radius form.

How do you identify the center and radius of a circle from its equation?

In the equation (x - h)^2 + (y - k)^2 = r^2, the values of h and k represent the x-coordinate and y-coordinate of the center, respectively. The value of r represents the radius of the circle.

Can the equation of a circle have a negative radius?

No, the radius of a circle cannot be negative. It represents the distance from the center of the circle to any point on its circumference, so it must always be a positive value.

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