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In the diagram, AE is tangent to the circle at point a and secant DE intersects the circle at point C and D. the iines intersect outside of the circle at point E

In the diagram, AE is tangent to the circle at point a and secant DE intersects the circle at point C and D. the iines intersect outside of the circle at point E
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in the diagram, AE is tangent to the circle at point a and secant DE intersects the circle at point C and D. the iines intersect outside of the circle at point E

User Dwergkees
by
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1 Answer

11 votes
Recall the secant-tangent theorem, and you have
EA^2 = EC*CD
12^2 = 8*(x+10)
and now ED = EC+CD = 8+x+10

I suspect a typo somewhere in the murk above
User Ahamed
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9.0k points
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