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\frac{-1}{7} [\latex]. The normal of this tangent will then have the negative reciprocal of this value. The gradient of the normal is [latex] \frac{dy}{dx} [/latex] which is [latex] \frac{7a}{a} [/latex]. This means the point at which the line [latex] y = \frac{c-x}{7} [/latex] meets the circle has coordinate ([3+a] , [4+7a]). Substituing into [br][br][latex] (x-3)^2 + (y-4)^2 = 25 [/latex] gives a = [late] \frac{\sqrt2}{2} [/late]
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Last reply 2 days ago
Did Cambridge maths students find maths and further maths a level very easy?Last reply 2 weeks ago
Edexcel A Level Mathematics Paper 2 unofficial mark scheme correct me if wrong71
Last reply 2 days ago
Did Cambridge maths students find maths and further maths a level very easy?Last reply 2 weeks ago
Edexcel A Level Mathematics Paper 2 unofficial mark scheme correct me if wrong71