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Circle problem finding coordinates of points

Casio

Member
Feb 11, 2012
86
Continued from;
Originally Posted by Jameson What is the full problem you are trying to solve? I can't make sense of your post until I know that.


I have a circle problem and am trying to find coordinates of any points at which the circle (x + 3)^2 + (y - 4)^2 = 17 intersects the line 3y = - 5x + 14.

I started off and got to;

(x + 3)^2 + ( y - 4)^2 = 17

(x + 3)^2 + ( - 5x + 14 - 4) = 17

(x + 3)^2 + ( - 5x + 2/3) = 17

(x + 3)^2 + 1/9(25x^2 - 20x + 4) = 17

I got this far above but don't know what to do with the denominator 9?

If I expand (x + 3)^2 = x^2 + 6x + 9

What I can't do is add this to 1/9(25x^2 - 20x + 4) = 17

This is were I am stuck?
 
Last edited:

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Why are you changing -5x + 14 - 4 to -5x + 2/3?

I would write:

$\displaystyle (x+3)^2+(y-4)^2=17$

Substitute for y:

$\displaystyle (x+3)^2+(-5x+14-4)^2=17$

Combine like terms within second term of equation, and factor out $\displaystyle -5$:

$\displaystyle (x+3)^2+(-5)^2(x-2)^2=17$

$\displaystyle (x+3)^2+25(x-2)^2=17$

Now, expand and write in standard form the resulting quadratic in x.
 

Casio

Member
Feb 11, 2012
86
Why are you changing -5x + 14 - 4 to -5x + 2/3?

I would write:

$\displaystyle (x+3)^2+(y-4)^2=17$

Substitute for y:

$\displaystyle (x+3)^2+(-5x+14-4)^2=17$

Combine like terms within second term of equation, and factor out $\displaystyle -5$:

$\displaystyle (x+3)^2+(-5)^2(x-2)^2=17$

$\displaystyle (x+3)^2+25(x-2)^2=17$

Now, expand and write in standard form the resulting quadratic in x.
Just a query that I don't understand how you got from;

$\displaystyle (x + 3)^2 + ( - 5x + 14 - 4)^2 = 17$

To

$\displaystyle (x + 3)^2 + 25( x - 2)^2 = 17$

Thanks

Mark you asked;
Why are you changing -5x + 14 - 4 to -5x + 2/3?

Because the circle intersects the line 3y = - 5x + 14

therefore - 5x + 14 is divided by 3 and the minus 4 became 2\3
 
Last edited:

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Oops! Sorry, I misread the problem. Let me try this again...

I would write:

$\displaystyle (x+3)^2+(y-4)^2=17$

Substitute for y:

$\displaystyle (x+3)^2+\left(\frac{-5x+14}{3}-4 \right)^2=17$

$\displaystyle (x+3)^2+\left(\frac{-5x+14-12}{3} \right)^2=17$

$\displaystyle (x+3)^2+\left(\frac{-5x+2}{3} \right)^2=17$

Using $\displaystyle (a-b)^2=(b-a)^2$ we may write:

$\displaystyle (x+3)^2+\left(\frac{5x-2}{3} \right)^2=17$

Factor out the square of 1/3:

$\displaystyle (x+3)^2+\frac{1}{9}(x-2)^2=17$

Multiply through by 9:

$\displaystyle 9(x+3)^2+(x-2)^2=153$

Now, expand, distribute and write in standard form the resulting quadratic in x.
 

soroban

Well-known member
Feb 2, 2012
409
Hello, everyone!

$\text{Find the intersections of the circle }\,(x + 3)^2 + (y - 4)^2 \:=\: 17\;\;\color{blue}{[1]}$
$\text{ and the line }\,3y \:=\: -5x + 14\;\;\color{blue}{[2]}$

From [2]: .$y \:=\:\text{-}\frac{5}{3}x + \frac{14}{3}$

Substitute into [1]: .$(x+3)^2 + \left(\text{-}\frac{5}{3}x+\frac{14}{3} - 4\right)^2 \;=\;17$

. . . . . . . . . . . . . . . . . $(x+3)^2 + \left(\text{-}\frac{5}{3}x + \frac{2}{3}\right)^2 \;=\;17$

. . . . . . . . . . . . $x^2 + 6x + 9 + \frac{25}{9}x^2 - \frac{20}{9}x + \frac{4}{9} \;=\;17$

Multiply by 9: .$9x^2 + 54x + 81 + 25x^2 - 20x + 4 \;=\;153$

. . . . . . . . . . . . . . . . . . . . . . .$34x^2 + 34x - 68 \;=\;0$

Divide by 34: .$x^2 + x - 2 \;=\;0$

. . . . . . . .$(x-1)(x+2) \;=\;0$

. . . . . . . . . . . $ x \:=\:1,\:\text{-}2$

. . . . . . . . . . . $ y \:=\:3,\:8$


Answers: .$(1,\,3),\;(\text{-}2,\,8)$
 

Casio

Member
Feb 11, 2012
86
Hello, everyone!


From [2]: .$y \:=\:\text{-}\frac{5}{3}x + \frac{14}{3}$

Substitute into [1]: .$(x+3)^2 + \left(\text{-}\frac{5}{3}x+\frac{14}{3} - 4\right)^2 \;=\;17$

. . . . . . . . . . . . . . . . . $(x+3)^2 + \left(\text{-}\frac{5}{3}x + \frac{2}{3}\right)^2 \;=\;17$

. . . . . . . . . . . . $x^2 + 6x + 9 + \frac{25}{9}x^2 - \frac{20}{9}x + \frac{4}{9} \;=\;17$

Multiply by 9: .$9x^2 + 54x + 81 + 25x^2 - 20x + 4 \;=\;153$

. . . . . . . . . . . . . . . . . . . . . . .$34x^2 + 34x - 68 \;=\;0$

Divide by 34: .$x^2 + x - 2 \;=\;0$

. . . . . . . .$(x-1)(x+2) \;=\;0$

. . . . . . . . . . . $ x \:=\:1,\:\text{-}2$

. . . . . . . . . . . $ y \:=\:3,\:8$


Answers: .$(1,\,3),\;(\text{-}2,\,8)$
Thanks, I was getting there but was struggling to understand initially how to get rid of the denominator 9, which I multiplied through like you and got the results, but I must admit at the point of 34x^2 + 34x - 68 = 0, I had difficulties because I was getting solutions saying I had no roots and the question said the line did intersect the circle.

I don't think I would have thought about dividing the 34's out and would not have solved this without your help.

Very much appreciated to all that contributed.