Solve for k k - 2/ 5 = 11j
2 answers:
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the correct question is
<span>What value of x would make RQ tangent to Circle P at point Q?
the picture in the attached figure
we know that
If RQ is a tangent to circle P at point Q
then
</span>RQ is perpendicular to the radius PQ at the point of tangency Q.
This means that ΔPRQ would be a right triangle.
Applying Pythagorean theorem
RP²=PQ²+RQ²
(x+9)²=9²+12²
(x+9)²=225
(+/-)(x+9)=√225
(+/-)(x+9)=15
+(x+9)=15-----> x=15-9-----> x=6
-(x+9)=15-----> -x-9=15-----> x=-9-15-----> x=-24
the solution is
x=6
the answer is
x=6 units
<span>What value of x would make RQ tangent to Circle P at point Q?
the picture in the attached figure
we know that
If RQ is a tangent to circle P at point Q
then
</span>RQ is perpendicular to the radius PQ at the point of tangency Q.
This means that ΔPRQ would be a right triangle.
Applying Pythagorean theorem
RP²=PQ²+RQ²
(x+9)²=9²+12²
(x+9)²=225
(+/-)(x+9)=√225
(+/-)(x+9)=15
+(x+9)=15-----> x=15-9-----> x=6
-(x+9)=15-----> -x-9=15-----> x=-9-15-----> x=-24
the solution is
x=6
the answer is
x=6 units