@@jooniespie It looks like you are trying to get a perfect SAT score. Very few people would be interested in solving this problem algebraically, specially with Desmos. a = 5 does not make any sense. It has radius 2.5 and center (2.5, -1). You can draw the circle. It does not intersect the first circle at one point. (a/2 - 1/2)^2 + (1)^2 = (a/2 + 1/2)^2 (a - 1)^2 - (a + 1)^2 = - 4 - 4a = -4 a = 1 I am doing this for fun only. Do not be surprised.
why are u assuming the intersctioon should be at the same value of x of each circles just bcs of one lined? like it could be anywhere for example when it's sloping
@@britishYES Watch the Desmos simulation at 0:50. The second circle is not going to intersect the first circle at an angle. In order to fully understand this, you need to study how “a” affects the given equation (size and positioning).
This is really easy to solve with Desmos. Just plug both of the equations into Desmos and use a slider for "a". Given that "a" is greater than -5 but less than 10, it narrows down the values of what "a" could be.
![[December SAT] 4 Must Know Geometry Skills (2024 Digital)](/img/1.gif)
this is the perfect example to use desmos
Hey do you know which subreddit this question is from or do you know any subreddits that post harder questions? Thanks!
This is really good. Helped a lot.
could you please explain the algebraic way of solving it too?
The distance between the centres should equal the sum of the circles radii. This should give you an equation to solve for a
@@DrewWerbowskigot it, thanks!! im getting two solutions though, 1 and 5, both of them are within the lim too
@@jooniespie
It looks like you are trying to get a perfect SAT score. Very few people would be interested in solving this problem algebraically, specially with Desmos.
a = 5 does not make any sense. It has radius 2.5 and center (2.5, -1). You can draw the circle. It does not intersect the first circle at one point.
(a/2 - 1/2)^2 + (1)^2 = (a/2 + 1/2)^2
(a - 1)^2 - (a + 1)^2 = - 4
- 4a = -4
a = 1
I am doing this for fun only. Do not be surprised.
why are u assuming the intersctioon should be at the same value of x of each circles just bcs of one lined? like it could be anywhere for example when it's sloping
Not assuming, just noticing that is a possible solution and should be easier to solve than if its centre were at a different x-value
@@DrewWerbowski hmm thx for letting me answered
@@britishYES
Watch the Desmos simulation at 0:50. The second circle is not going to intersect the first circle at an angle. In order to fully understand this, you need to study how “a” affects the given equation (size and positioning).
We can simply get 1 by comparing with each other .
could you pls explain how
What if the second equation is
x^2 + y^2 - ax + 3y + 1 = 0
This is really easy to solve with Desmos. Just plug both of the equations into Desmos and use a slider for "a". Given that "a" is greater than -5 but less than 10, it narrows down the values of what "a" could be.
Yep, addressed this in the first part of the video.