1. ## Can you help me?(Question about circle)

The circle S has equation :

X2+Y2- 2X-4Y=8
Show thatA(-1,-1)and B (3,5) are ends of diameter

Show that C(4,4)lies on S and determine angle ACB

Find the distance from D(5,0) to the nearest point E on S

Find the the x-coordinates of the points of intersection of S with the line joining D and F(0,2).

2. Originally Posted by Diligent_Learner
The circle S has equation :

$x^2+y^2- 2x-4y=8$

Show thatA(-1,-1)and B (3,5) are ends of diameter

Complete the square in the equation for both x and y to locate the center. Then, find the midpoint of AB to see if they are one and the same. The center of the circle would be the midpoint of the diameter.

$(x^2-2x+1)+(y^2-4y+4)=8+5$

$(x-1)^2+(y-2)^2=13$

Center is at (1, 2)

Midpoint of AB = $\left(\frac{3-1}{2}, \ \ \frac{5-1}{2}\right)={\color{red}(1, 2)}$

Originally Posted by Diligent_Learner
Show that C(4,4)lies on S and determine angle ACB

Substitute (4, 4) into your original equation to see if it makes a true statement

$4^2+4^2- 2(4)-4(4)=8$??

Angle ACB is an inscribed angle in a semicircle, therefore 90 degrees.

Originally Posted by Diligent_Learner
Find the distance from D(5,0) to the nearest point E on S

See Earboth's post.

Originally Posted by Diligent_Learner
Find the the x-coordinates of the points of intersection of S with the line joining D and F(0,2).
Define a linear equation for the line passing through D(5, 0) and F(0, 2). Then solve the system which includes this linear equation and the equaton of the circle. You should get two distinct points of intersection. Try it.

3. Thank you very much for your great help.I understand the question

The equation of the line that passes through D and F is:

y-y1=m(x-x1)

Y= -0.4x+2

====================

X2+y2-2x-4y=8

X2+(-0.4x+2)2-2x-4(-0.4x+2)

X2-3.6X-12=0

X=5.7 or X=-2.1

4. Originally Posted by Diligent_Learner
[FONT=Times New Roman]...

Find the distance from D(5,0) to the nearest point E on S
...
Let C denote the center of the circle. Then the Point E lies on the circle and on the line CD:

$CD: y =-\dfrac12 x + \dfrac52$

Calculating the intersection points of the circle and this line:

$(x-1)^2+\left(-\dfrac12 x + \dfrac52 - 2\right)^2 = 13$

yields: $x = 1\pm \frac15 \sqrt{65}$ The x-coordinate of E must be greater than 1: Plug in this value into the equation of the line to get the y-coordinate. I've got $E\left(1+\frac15\sqrt{65}\ ,\ 2-\frac15\sqrt{65}\right)$

The distance $ED = CD - r~\implies~ ED=\sqrt{20} - \sqrt{13} \approx 0.86658...$

5. Thank you very much for clarifying the idea ..

and Can yiu tell me about if my answer is true in the previous post?

and in which program you draw the circle?

Originally Posted by Diligent_Learner
Thank you very much for clarifying the idea ..

and Can yiu tell me about if my answer is true in the previous post?

and in which program you draw the circle?
You've got the equation of the line correctly.
But I've got results, different from yours, calculating the points of intersection:

$l:y=-\dfrac25 x+2$

Plug in the term of y into the equation of the circle:

$(x-1)^2+\left(-\dfrac25 x+2 - 2\right)^2=13$

$x^2-2x+1+\dfrac4{25}x^2=13$

$\dfrac{29}{25} x^2 - 2x-12=0$

Use the quadratic formula to solve this equation:

$x=\dfrac{25}{29} \pm \dfrac5{29} \sqrt{373}$

Take these x-values and plug them into the equation of the line to get the y-coordinates of the points of intersection.

EDIT: Use the attached sketch to control your results.