Two tangents drawn to a circle from a point outside it, are equal in length.prove it.
1 answer:
Given: A circle with centre O; PA and PB are two tangents to the circle drawn from an external point P.
To prove: PA = PB
Construction: Join OA, OB, and OP.
It is known that a tangent at any point of a circle is perpendicular to the radius through the point of contact.
OA⊥PA
OB⊥PB
In △OPA and △OPB
∠OPA=∠OPB (Using (1))
OA=OB (Radii of the same circle)
OP=OP (Common side)
Therefor △OPA≅△OPB (RHS congruency criterion)
PA=PB
(Corresponding parts of congruent triangles are equal)
Thus, it is proved that the lengths of the two tangents drawn from an external point to a circle are equal.
The length of tangents drawn from any external point are equal.
So statement is correct
To prove: PA = PB
Construction: Join OA, OB, and OP.
It is known that a tangent at any point of a circle is perpendicular to the radius through the point of contact.
OA⊥PA
OB⊥PB
In △OPA and △OPB
∠OPA=∠OPB (Using (1))
OA=OB (Radii of the same circle)
OP=OP (Common side)
Therefor △OPA≅△OPB (RHS congruency criterion)
PA=PB
(Corresponding parts of congruent triangles are equal)
Thus, it is proved that the lengths of the two tangents drawn from an external point to a circle are equal.
The length of tangents drawn from any external point are equal.
So statement is correct
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Answer:
see attached
Step-by-step explanation:
I find it convenient to let a graphing calculator draw the graph (attached).
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If you're drawing the graph by hand, there are a couple of strategies that can be useful.
The first equation is almost in slope-intercept form. Dividing it by 2 will put it in that form:
y = 2x -4
This tells you that the y-intercept, (0, -4) is a point on the graph, as is the point that is up 2 and right 1 from there: (1, -2). A line through those points completes the graph.
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The second equation is in standard form, so the x- and y-intercepts are easily found. One way to do that is to divide by the constant on the right to get ...
x/2 +y/3 = 1
The denominators of the x-term and the y-term are the x-intercept and the y-intercept, respectively. If that is too mind-bending, you can simply set x=0 to find the y-intercept:
0 +2y = 6
y = 6/2 = 3
and set y=0 to find the x-intercept
3x +0 = 6
x = 6/3 = 2
Plot the intercepts and draw the line through them for the graph of this equation.
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Here, we have suggested graphing strategies that don't involve a lot of manipulation of the equations. The idea is to get there as quickly as possible with a minimum of mistakes.
