Find the Length of $QP$Help: Find the area of the shaded regionThree mutually-tangent circles have centers at given distances from each other; find each radius, and find the area between the circlesProve that the length of the common chord is $frac2absin thetasqrta^2+b^2+2abcos theta$Show that the circles touch externally and find the coordinate where thy touchLocus of point of intersection of tangents at $A$ and $B$Finding vertices of rhombus formed by lines $y=2x+4$, $y=-frac13x+4$ and $(12,0)$ is a vertex. Can't find last vertex.Equation of circles tangent to x-axis with a given radius and pointProbability that length of Randomly chosen chord of a circleangle between two circles(which angle to be considered?)In $triangle ABC$ with $D$ on $overlineAC$, if $angle CBD=2angle ABD$ and the circumcenter lies on $overlineBC$, then $AD/DCneq 1/2$
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Find the Length of $QP$
Help: Find the area of the shaded regionThree mutually-tangent circles have centers at given distances from each other; find each radius, and find the area between the circlesProve that the length of the common chord is $frac2absin thetasqrta^2+b^2+2abcos theta$Show that the circles touch externally and find the coordinate where thy touchLocus of point of intersection of tangents at $A$ and $B$Finding vertices of rhombus formed by lines $y=2x+4$, $y=-frac13x+4$ and $(12,0)$ is a vertex. Can't find last vertex.Equation of circles tangent to x-axis with a given radius and pointProbability that length of Randomly chosen chord of a circleangle between two circles(which angle to be considered?)In $triangle ABC$ with $D$ on $overlineAC$, if $angle CBD=2angle ABD$ and the circumcenter lies on $overlineBC$, then $AD/DCneq 1/2$
$begingroup$
Given two circles with radii $8$ and $6$ units with centers $A$ and $B$ such that $AB=12$ If $P$ is mid point of $QR$
Find Length of $QP$
My try:
I assumed $A(0,0)$ and $B(12,0)$
So the equations of circles are:
$$x^2+y^2=64$$
$$(x-12)^2+y^2=36$$
Solving above equations we get:
$$Pleft(frac436, fracsqrt4556right)$$
Then we can assume
$$Q(8cosalpha, 8sin alpha)$$
$$R(12+6cosbeta,6sinbeta)$$
Using mid point formula we get:
$$8cosalpha+6cosbeta=frac73$$
$$8sinalpha+6sinbeta=fracsqrt4553$$
Any idea here?
euclidean-geometry analytic-geometry circle
asked Mar 11 at 10:39
Umesh shankarUmesh shankar
2,99331220
$endgroup$
add a comment |
$begingroup$
Given two circles with radii $8$ and $6$ units with centers $A$ and $B$ such that $AB=12$ If $P$ is mid point of $QR$
Find Length of $QP$
My try:
I assumed $A(0,0)$ and $B(12,0)$
So the equations of circles are:
$$x^2+y^2=64$$
$$(x-12)^2+y^2=36$$
Solving above equations we get:
$$Pleft(frac436, fracsqrt4556right)$$
Then we can assume
$$Q(8cosalpha, 8sin alpha)$$
$$R(12+6cosbeta,6sinbeta)$$
Using mid point formula we get:
$$8cosalpha+6cosbeta=frac73$$
$$8sinalpha+6sinbeta=fracsqrt4553$$
Any idea here?
euclidean-geometry analytic-geometry circle
asked Mar 11 at 10:39
Umesh shankarUmesh shankar
2,99331220
$endgroup$
$begingroup$
solve for $cos alpha$ and $sin alpha$ from the last 2 equations
$endgroup$
– Soumik Ghosh
Mar 11 at 11:49
add a comment |
$begingroup$
Given two circles with radii $8$ and $6$ units with centers $A$ and $B$ such that $AB=12$ If $P$ is mid point of $QR$
Find Length of $QP$
My try:
I assumed $A(0,0)$ and $B(12,0)$
So the equations of circles are:
$$x^2+y^2=64$$
$$(x-12)^2+y^2=36$$
Solving above equations we get:
$$Pleft(frac436, fracsqrt4556right)$$
Then we can assume
$$Q(8cosalpha, 8sin alpha)$$
$$R(12+6cosbeta,6sinbeta)$$
Using mid point formula we get:
$$8cosalpha+6cosbeta=frac73$$
$$8sinalpha+6sinbeta=fracsqrt4553$$
Any idea here?
euclidean-geometry analytic-geometry circle
asked Mar 11 at 10:39
Umesh shankarUmesh shankar
2,99331220
$endgroup$
Given two circles with radii $8$ and $6$ units with centers $A$ and $B$ such that $AB=12$ If $P$ is mid point of $QR$
Find Length of $QP$
My try:
I assumed $A(0,0)$ and $B(12,0)$
So the equations of circles are:
$$x^2+y^2=64$$
$$(x-12)^2+y^2=36$$
Solving above equations we get:
$$Pleft(frac436, fracsqrt4556right)$$
Then we can assume
$$Q(8cosalpha, 8sin alpha)$$
$$R(12+6cosbeta,6sinbeta)$$
Using mid point formula we get:
$$8cosalpha+6cosbeta=frac73$$
$$8sinalpha+6sinbeta=fracsqrt4553$$
Any idea here?
euclidean-geometry analytic-geometry circle
euclidean-geometry analytic-geometry circle
asked Mar 11 at 10:39
Umesh shankarUmesh shankar
2,99331220
asked Mar 11 at 10:39
Umesh shankarUmesh shankar
2,99331220
asked Mar 11 at 10:39
Umesh shankarUmesh shankar
2,99331220
asked Mar 11 at 10:39
Umesh shankarUmesh shankar
2,99331220
asked Mar 11 at 10:39
Umesh shankarUmesh shankar
2,99331220
2,99331220
$begingroup$
solve for $cos alpha$ and $sin alpha$ from the last 2 equations
$endgroup$
– Soumik Ghosh
Mar 11 at 11:49
add a comment |
$begingroup$
solve for $cos alpha$ and $sin alpha$ from the last 2 equations
$endgroup$
– Soumik Ghosh
Mar 11 at 11:49
$begingroup$
solve for $cos alpha$ and $sin alpha$ from the last 2 equations
$endgroup$
– Soumik Ghosh
Mar 11 at 11:49
$begingroup$
solve for $cos alpha$ and $sin alpha$ from the last 2 equations
$endgroup$
– Soumik Ghosh
Mar 11 at 11:49
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You don't need any trigonometry or analytical geometry. The following equations are quite obvious (Pythagora):
$$p^2+fracx^24=8^2=64tag1$$
$$q^2+fracx^24=6^2=36tag2$$
$$(p-q)^2+x^2=12^2=144tag3$$
Introduce substitution: $y=fracx^24$ and you get:
$$p=sqrt64-ytag4$$
$$q=sqrt36-ytag5$$
$$(p-q)^2+4y=144tag6$$
Replace (4) and (5) into (6):
$$(sqrt64-y-sqrt36-y)^2+4y=144$$
$$100-2y-2sqrt(64-y)(36-y)+4y=144$$
$$2y-44 = 2sqrt(64-y)(36-y)$$
$$y-22=sqrt(64-y)(36-y)$$
$$(y-22)^2=(64-y)(36-y)$$
$$(y-22)^2-(64-y)(36-y)=0$$
$$56y-1820=0$$
$$y=fracx^24=frac182056=frac652$$
$$x=sqrt130$$
answered Mar 11 at 12:44
OldboyOldboy
8,67911036
$endgroup$
$begingroup$
Are the blue dots centers of the circles?
$endgroup$
– Umesh shankar
Mar 11 at 13:02
$begingroup$
@Umeshshankar Yes, they are.
$endgroup$
– Oldboy
Mar 11 at 13:13
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You don't need any trigonometry or analytical geometry. The following equations are quite obvious (Pythagora):
$$p^2+fracx^24=8^2=64tag1$$
$$q^2+fracx^24=6^2=36tag2$$
$$(p-q)^2+x^2=12^2=144tag3$$
Introduce substitution: $y=fracx^24$ and you get:
$$p=sqrt64-ytag4$$
$$q=sqrt36-ytag5$$
$$(p-q)^2+4y=144tag6$$
Replace (4) and (5) into (6):
$$(sqrt64-y-sqrt36-y)^2+4y=144$$
$$100-2y-2sqrt(64-y)(36-y)+4y=144$$
$$2y-44 = 2sqrt(64-y)(36-y)$$
$$y-22=sqrt(64-y)(36-y)$$
$$(y-22)^2=(64-y)(36-y)$$
$$(y-22)^2-(64-y)(36-y)=0$$
$$56y-1820=0$$
$$y=fracx^24=frac182056=frac652$$
$$x=sqrt130$$
answered Mar 11 at 12:44
OldboyOldboy
8,67911036
$endgroup$
$begingroup$
Are the blue dots centers of the circles?
$endgroup$
– Umesh shankar
Mar 11 at 13:02
$begingroup$
@Umeshshankar Yes, they are.
$endgroup$
– Oldboy
Mar 11 at 13:13
add a comment |
$begingroup$
You don't need any trigonometry or analytical geometry. The following equations are quite obvious (Pythagora):
$$p^2+fracx^24=8^2=64tag1$$
$$q^2+fracx^24=6^2=36tag2$$
$$(p-q)^2+x^2=12^2=144tag3$$
Introduce substitution: $y=fracx^24$ and you get:
$$p=sqrt64-ytag4$$
$$q=sqrt36-ytag5$$
$$(p-q)^2+4y=144tag6$$
Replace (4) and (5) into (6):
$$(sqrt64-y-sqrt36-y)^2+4y=144$$
$$100-2y-2sqrt(64-y)(36-y)+4y=144$$
$$2y-44 = 2sqrt(64-y)(36-y)$$
$$y-22=sqrt(64-y)(36-y)$$
$$(y-22)^2=(64-y)(36-y)$$
$$(y-22)^2-(64-y)(36-y)=0$$
$$56y-1820=0$$
$$y=fracx^24=frac182056=frac652$$
$$x=sqrt130$$
answered Mar 11 at 12:44
OldboyOldboy
8,67911036
$endgroup$
$begingroup$
Are the blue dots centers of the circles?
$endgroup$
– Umesh shankar
Mar 11 at 13:02
$begingroup$
@Umeshshankar Yes, they are.
$endgroup$
– Oldboy
Mar 11 at 13:13
add a comment |
$begingroup$
You don't need any trigonometry or analytical geometry. The following equations are quite obvious (Pythagora):
$$p^2+fracx^24=8^2=64tag1$$
$$q^2+fracx^24=6^2=36tag2$$
$$(p-q)^2+x^2=12^2=144tag3$$
Introduce substitution: $y=fracx^24$ and you get:
$$p=sqrt64-ytag4$$
$$q=sqrt36-ytag5$$
$$(p-q)^2+4y=144tag6$$
Replace (4) and (5) into (6):
$$(sqrt64-y-sqrt36-y)^2+4y=144$$
$$100-2y-2sqrt(64-y)(36-y)+4y=144$$
$$2y-44 = 2sqrt(64-y)(36-y)$$
$$y-22=sqrt(64-y)(36-y)$$
$$(y-22)^2=(64-y)(36-y)$$
$$(y-22)^2-(64-y)(36-y)=0$$
$$56y-1820=0$$
$$y=fracx^24=frac182056=frac652$$
$$x=sqrt130$$
answered Mar 11 at 12:44
OldboyOldboy
8,67911036
$endgroup$
You don't need any trigonometry or analytical geometry. The following equations are quite obvious (Pythagora):
$$p^2+fracx^24=8^2=64tag1$$
$$q^2+fracx^24=6^2=36tag2$$
$$(p-q)^2+x^2=12^2=144tag3$$
Introduce substitution: $y=fracx^24$ and you get:
$$p=sqrt64-ytag4$$
$$q=sqrt36-ytag5$$
$$(p-q)^2+4y=144tag6$$
Replace (4) and (5) into (6):
$$(sqrt64-y-sqrt36-y)^2+4y=144$$
$$100-2y-2sqrt(64-y)(36-y)+4y=144$$
$$2y-44 = 2sqrt(64-y)(36-y)$$
$$y-22=sqrt(64-y)(36-y)$$
$$(y-22)^2=(64-y)(36-y)$$
$$(y-22)^2-(64-y)(36-y)=0$$
$$56y-1820=0$$
$$y=fracx^24=frac182056=frac652$$
$$x=sqrt130$$
answered Mar 11 at 12:44
OldboyOldboy
8,67911036
answered Mar 11 at 12:44
OldboyOldboy
8,67911036
answered Mar 11 at 12:44
OldboyOldboy
8,67911036
answered Mar 11 at 12:44
OldboyOldboy
8,67911036
8,67911036
$begingroup$
Are the blue dots centers of the circles?
$endgroup$
– Umesh shankar
Mar 11 at 13:02
$begingroup$
@Umeshshankar Yes, they are.
$endgroup$
– Oldboy
Mar 11 at 13:13
add a comment |
$begingroup$
Are the blue dots centers of the circles?
$endgroup$
– Umesh shankar
Mar 11 at 13:02
$begingroup$
@Umeshshankar Yes, they are.
$endgroup$
– Oldboy
Mar 11 at 13:13
$begingroup$
Are the blue dots centers of the circles?
$endgroup$
– Umesh shankar
Mar 11 at 13:02
$begingroup$
Are the blue dots centers of the circles?
$endgroup$
– Umesh shankar
Mar 11 at 13:02
$begingroup$
@Umeshshankar Yes, they are.
$endgroup$
– Oldboy
Mar 11 at 13:13
$begingroup$
@Umeshshankar Yes, they are.
$endgroup$
– Oldboy
Mar 11 at 13:13
add a comment |
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$begingroup$
solve for $cos alpha$ and $sin alpha$ from the last 2 equations
$endgroup$
– Soumik Ghosh
Mar 11 at 11:49