
Let AB be a chord of the given circle with centre and radius 13 cm.
Then, OA = 13 cm and ab = 10 cm
From O, draw OL⊥ AB
We know that the perpendicular from the centre of a circle to a chord bisects the chord.
∴ AL = ½AB = (½ × 10)cm = 5 cm
From the right △OLA, we have
OA² = OL² + AL²
==> OL² = OA² – AL²
==> [(13)² – (5)²] cm² = 144cm²
==> OL = √144cm = 12 cm
Hence, the distance of the chord from the centre is 12 cm.
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the parallel line is 2x+5y+15=0.
Step-by-step explanation:
ok I hope it will work
soo,
Solution
given,
given parallel line 2x+5y=15
which goes through the point (-10,1)
now,
let 2x+5y=15 be equation no.1
then the line which is parallel to the equation 1st
2 x+5y+k = 0 let it be equation no.2
now the equation no.2 passes through the point (-10,1)
or, 2x+5y+k =0
or, 2*-10+5*1+k= 0
or, -20+5+k= 0
or, -15+k= 0
or, k= 15
putting the value of k in equation no.2 we get,
or, 2x+5y+k=0
or, 2x+5y+15=0
which is a required line.
Answer:
The awnser is 2 and 3!!!!!!
Answer:
6x2-2x
Step-by-step explanation: