http://img8.imageshack.us/i/snapshot20090605.jpg/
for (i) i understand 2 figures are similar if they have same angles and length.
but i can only see common angle D, rest i cant visualize
(ii) from tangent-secant theorem,
DB^2= DC* DA
then how to continue to link to the proof?
help appreciated, thanks!
next question:
given equation of a circle is x^2 + y^2 -4x +8y - 16 = 0, find the
(i) coordinates of centre and radius of circle
(ii) equation of another circle such that the new circle is a reflection of
x^2 + y^2 -4x +8y - 16 = 0 in the y-axis.
i can solve (a) radius=6cm, centre (2,-4) but stuck at (b)
the new coordinates of centre of circle is now (-2,-4) after being reflected in y axis
so i wrote (x+2) ^2 + (y+4)^2 = 6 ( correct?)
Originally posted by Divanhot:http://img8.imageshack.us/i/snapshot20090605.jpg/
for (i) i understand 2 figures are similar if they have same angles and length.
but i can only see common angle D, rest i cant visualize
(ii) from tangent-secant theorem,
DB^2= DC* DA
then how to continue to link to the proof?
help appreciated, thanks!
(i)
-AB bisects angle DBC, hence angle ABC = angle ABD = 45degree
-angle ADB = angle CAB
-Both triangles have DB as a common side.
= > 2 same angles, and a common side, triangle ABD and BCD are similar triangles.
(ii)
DB^2 - DA^2 = DC*DA - DA*DA
=DA ( DC - DA )
=DA*AC
=AD*AB
Note that AC = AB.
Originally posted by Divanhot:next question:
given equation of a circle is x^2 + y^2 -4x +8y - 16 = 0, find the
(i) coordinates of centre and radius of circle
(ii) equation of another circle such that the new circle is a reflection of
x^2 + y^2 -4x +8y - 16 = 0 in the y-axis.
i can solve (a) radius=6cm, centre (2,4) but stuck at (b)
the new coordinates of centre of circle is now (-2,4) after being reflected in y axis
so i wrote (x+2) ^2 + (y-4)^2 = 6 ( correct?)
(x+2)^2+(y-4)^2=36
hi bro forbiddensinner, answer to the circle question part ii) should it be
(x+2)^2 + (y+4) ^2 = 6 ?
since coordinate of circle after reflection in y axis is (-2,-4)
hi bro forbidden, for the answer to proving DB^ 2- DA^2= AD.AB
i like to ask how you get AC= AB?
thxs
Originally posted by Divanhot:hi bro forbiddensinner, answer to the circle question part ii) should it be
(x+2)^2 + (y+4) ^2 = 6 ?
since coordinate of circle after reflection in y axis is (-2,-4)
(x-a)^2 + (y-b)^2 = r^2
You seem to have forgotten something, hehehe.....
Originally posted by Divanhot:hi bro forbidden, for the answer to proving DB^ 2- DA^2= AD.AB
i like to ask how you get AC= AB?
thxs
DBI is a tangent to the circle at B
=> BC is the diameter.
angle ABC= 45 degrees, and both AC and AB meet at A, which is on the circle.
=> AB = AC
hi bro, i understand now AB is the diameter of the circle and angle ABC= 45 degrees.
but i cant c if AC and AB meet at A, then AB=AC.. do guide me, i am lost!
Originally posted by Divanhot:hi bro, i understand now AB is the diameter of the circle and angle ABC= 45 degrees.
but i cant c if AC and AB meet at A, then AB=AC.. do guide me, i am lost!
Triangle ABC is an isoceles triangle.
angle ABC = 45 degrees, BC is the diameter, and A is on the circle.
Hence angle ACB must also be 45 degrees and AC = AB.
Maybe you will like to draw it out and try for youself. Or maybe our kind mod will help explain it more clearly.
