Parameter k, circle equation

A

Ana.stasia

Junior Member
Joined
Sep 28, 2020
Messages
118
Determine the parameter k so that the circle x2 + y2- (k-4) x - ky + 2k +5 = 0 touches the y axis.

So I thought since the circle touches the y axis I could use a point A (0,y), however, I didn't make any process from that.

IMG_20210313_215359.jpg
 
Ana.stasia said:
Determine the parameter k so that the circle x2 + y2- (k-4) x - ky + 2k +5 = 0 touches the y axis.

So I thought since the circle touches the y axis I could use a point A (0,y), however, I didn't make any process from that.

View attachment 25698
Click to expand...
You have started correctly.

You are trying to satisfy one condition - where the circle passes through 'y'. But the circle is touching the y-axis, NOT intersecting the y-axis.

What would be the condition for that?
 
Subhotosh Khan said:
You have started correctly.

You are trying to satisfy one condition - where the circle passes through 'y'. But the circle is touching the y-axis, NOT intersecting the y-axis.

What would be the condition for that?
Click to expand...

Well we have that formula r2(1+k2) = (kp-q+n)2 which is correct if a line is tangeny yo the circle. Can that be used?
 
Ana.stasia said:
Determine the parameter k so that the circle x2 + y2- (k-4) x - ky + 2k +5 = 0 touches the y axis.

So I thought since the circle touches the y axis I could use a point A (0,y), however, I didn't make any process from that.

View attachment 25698
Click to expand...
To be TANGENT - how many values of 'y' should you get?
 
[Determine the parameter k so that the circle x2 + y2- (k-4) x - ky + 2k +5 = 0 touches the y axis.
Click to expand...

at [MATH](0,y)[/MATH] ...

[MATH]y^2-ky+(2k+5) = 0[/MATH]
single solution for [MATH]y \implies b^2-4ac = (-k)^2 - 4(1)(2k+5) = 0[/MATH]
[MATH]k^2 - 8k -20 =(k-10)(k+2) =0 \implies k \in \{-2,10 \}[/MATH]
[MATH]k = -2 \implies x^2+6x+y^2+2y = -1[/MATH]
[MATH](x^2+6x+9)+(y^2+2y+1) = -1+9+1[/MATH]
[MATH](x+3)^2 + (y+1)^2 = 3^2[/MATH]
[MATH]k = 10 \implies x^2-6x+y^2-10y = -25[/MATH]
[MATH](x^2-6x+9) + (y^2-10y+25) = -25+9+25[/MATH]
[MATH](x-3)^2 + (y-5)^2 = 3^2[/MATH]
 
@skeeter What right did you have to put the whole constant (2k+5) with the y terms. I'm not saying that you are wrong but rather wondering why you are correct.
 
Jomo said:
@skeeter What right did you have to put the whole constant (2k+5) with the y terms. I'm not saying that you are wrong but rather wondering why you are correct.
Click to expand...
because if x = 0, all the non-zero terms form a quadratic in y ... ?‍♂️

take, for example, the circle [MATH](x-h)^2 + (y-k)^2 = r^2[/MATH] where [MATH](h,k)[/MATH] is in quadrant I.

[MATH]x = 0 \implies h^2 + (y-k)^2 = r^2 \implies (y-k)^2 = r^2-h^2 \implies y = k \pm \sqrt{r^2-h^2}[/MATH]
case 1 ... [MATH]r > h \implies \text{ two y-intercepts at } (0, k+\sqrt{r^2-h^2}) \text{ and } (0, k-\sqrt{r^2-h^2})[/MATH]
case 2 ... [MATH]r = h \implies \text{ one y-intercept at } (0, k)[/MATH]
case 3 ... [MATH]r < h \implies \text{ no y-intercepts because } r^2 < h^2 \implies \sqrt{r^2-h^2} \text{ is not a real value}[/MATH]
 
You must log in or register to reply here.
Top