## anonymous 5 years ago i need help with geometry

1. anonymous

$79\sin 5+3$

2. anonymous

Hi wgrac, what do you need to find out about what you wrote?

3. anonymous

$\sqrt{(-2-4)^{2}+(-3-y)^{2}}=10$

4. anonymous

Get rid of the radical and clean up your numbers. So, first, notice (-2-4)^2=(-6)^2=36. Also, $(-3-y)^2=(-(3+y))^2=(-1)^2(3+y)^2=(3+y)^2$So,$\sqrt{(-2-4)^2+(-3-y)^2}=10 \rightarrow 36+(3+y)^2=100$Subtract 36 from both sides$36+(3+y)^2=100 \rightarrow (3+y)^2=64 \rightarrow 3+y = \pm \sqrt{64}=\pm 8$That is,$3+y=\pm8 \rightarrow y = -3 \pm 8$

5. anonymous

So y=5 or y=-11.

6. anonymous

given equation of circles $x ^{2}+y ^{2}=2$ and $(x-3)^{2}+(y-3)^{2}=32$ explain why the circles must be internally tangent.

7. anonymous

given equation of circles $x ^{2}+y ^{2}=2$ and $(x-3)^{2}+(y-3)^{2}=32$ explain why the circles must be internally tangent.

8. anonymous

given equation of circles x2+y2=2 and (x−3)2+(y−3)2=32 explain why the circles must be internally tangent.

9. anonymous

Hello help, Two circles in the same plane are internally tangent if they intersect in exactly one point and the intersection of their interiors is not empty. Look at the attachment of the geometry of your problem. The smaller circle (x^2+y^2=2) is inside the larger one, and they intersect at one point. You have to do two things to show internal tangency. 1) Show that the circles touch at one point only 2) Show that the smaller circle lies inside, not outside, the larger circle. Okay, to answer part 1, we have to find the points of intersection of the two circles. Since we had a look at the geometry of the situation beforehand, we're expecting the circles to intersect at only one point. So, finding your points of intersection is a matter of solving simultaneous equations: $x^2+y^2=2$and$(x-3)^2+(y-3)^2=32$Take the first equation and solve for y:$y=\pm \sqrt{2-x^2}$We should test this by substituting into your second equation. If you take the positive square root first and substitute into the second equation, you get$(x-3)^2+(\pm \sqrt{2-x^2}-3)^2=32$which boils down to solving$\sqrt{2-x^2} \pm (x+2)=0$that is$\sqrt{2-x^2}=\mp (x+2) \rightarrow 2-x^2=(x+2)^2$Expand the right-hand side, collect your terms and solve the quadratic to get$x=-1$There is only one solution to this equation, even though we've considered both possible y's. So let's find the y-value that goes with this x-value on our circles. You can choose either of the initial equaations to start, so choose the simplest one,$(-1)^2+y^2=2 \rightarrow y^2=1 \rightarrow y=\pm 1$Now, although any one of these two possibilities satisfies the equation we just used, only y=-1 will satisfy the second circle equation as well when x=-1:$((-1)-3)^2+((1)-3)^2=20 \neq32$$((-1)-3)^2+((-1)-3)^2=(-4)^2+(-4)^2=32$So your point of intersection is (-1,-1) (there is only one). End of first part.

10. anonymous

Now you need to show that the smaller circle (one with smaller radius) lies entirely inside the larger. If one circle touches another circle at only one point, it must be the case that ALL of the circle is either inside the larger circle, or outside. If this wasn't the case, the smaller circle would cut the larger circle in more than one place (i.e. two) and part of the smaller circle would be inside the larger circle, and part of it outside. You can see this for yourself by drawing two circles that intersect at two points. All you have to do is pick a point other than (-1,-1) on your small circle, and show that the distance between that point and the center of the large circle is LESS THAN the radius of the large circle (which is sqrt{32}=4sqt{2}).

11. anonymous

So, you know that the point (1,1) is on the smaller circle since the left-hand side of the small circle's equation gives, $x^2+y^2=1^2+1^2$which is indeed equal to 2, the right-hand side. Now, if the distance between this point and the center of the larger circle is less than the radius of the larger circle, since we only have one point of intersection, it must be the case that ALL other points EITHER lie inside or outside the circle. If this point, (1,1) is inside, then so are all the others. We have for the distance, $d^2=(1-3)^2+(1-3)^2=8 \rightarrow d=\sqrt{8}=2\sqrt{2}$But the radius of the larger circle is $4\sqrt{2}$Since $2\sqrt{2}<4\sqrt{2}$it must be that (1,1) lies INSIDE the larger circle. Therefore, the entire small circle lies inside the larger circle. Since the smaller circle touches the larger one at one point only, and the smaller circle lies inside the larger, it must be that the circles are internally tangent.

12. anonymous

13. anonymous

14. anonymous

i forgot how slopes work.. a line with slope 3/4 passes through points (2,3) and (10,?)

15. anonymous

Not at a computer at the moment and difficult to write solution on phone. You can make a new post and someone else can take your problem. I can help later, but might be several hours - don't know if you want to wait that long.

16. anonymous

17. anonymous

prove that the segment joining the midpoints of the diags. of a trapezoid is parallel to the bases and has a length equal to half the difference of the lengths of the bases.

18. anonymous

& from the above problem, the vertices of the trap. is (a,0), (0,0), (b,c), and (d,c). its coordinate geom.

19. anonymous

20. anonymous

You're welcome, help.

21. anonymous

thanks, but when i tried using the coordinates given from my book. how come it didnt work?? aghhh. sorry, can you explain using the coordinates given to me? here is the diagram attached.

22. anonymous

23. anonymous

24. myininaya

hey lokisan can you tell me what i am missing in my problem. just follow me when and if you get a chance

25. anonymous

thankS a lot:)))

26. anonymous

ok np

27. anonymous

Suppose F is on PQ and PF=3/8PQ. If P=(x1,y1) and Q=(x2,y2), where x1<x2, find the coordinates of F.