A circle is centered at the point (-3, 2) and passes through the point (1, 5). The radius of the circle is how many units. The p
oint (-7,? ) lies on this circle.
2 answers:
We know that
the equation of a circle is
(x-h)²+(y-k)²=r²
(h,k) is the center----------> (-3,2)
the radius is the distance point (-3,2) and point <span>(1, 5)
d=</span>√(5-2)²+(1+3)²-------> d=√9+16----------> d=5
radius=5 units
equation is
(x+3)²+(y-2)²=25
for x=-7
(-7+3)²+(y-2)²=25-------> 16+y²-4y+4=25
y²-4y-5=0
using a graph tool -------> to resolve the second order equation
see the attached figure
the solution is
y1=-1
y2=5
the point will be
(-7,-1)
(-7,5)
the equation of a circle is
(x-h)²+(y-k)²=r²
(h,k) is the center----------> (-3,2)
the radius is the distance point (-3,2) and point <span>(1, 5)
d=</span>√(5-2)²+(1+3)²-------> d=√9+16----------> d=5
radius=5 units
equation is
(x+3)²+(y-2)²=25
for x=-7
(-7+3)²+(y-2)²=25-------> 16+y²-4y+4=25
y²-4y-5=0
using a graph tool -------> to resolve the second order equation
see the attached figure
the solution is
y1=-1
y2=5
the point will be
(-7,-1)
(-7,5)
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Answer:
BC:BN=8:3
Step-by-step explanation:
ABCD is a trapezoid and there is a point m which belongs to AD such that AM:MD=3:5.Line "l" parallel to AB intersects the diagonal AC at p and BD at N.
Now, we know that the parallel lines divide the transversal into the segments with equal ratio, therefore, BN:NC=AM:MD
But, BC= BN+NC
Therefore, BC:BN=(BN+NC):BN
⇒BC:BN=(3+5):3
⇒BC:BN=8:3
