The point $$\(A\)$$ has coordinates $$\((1,5)\)$$ and the line $$\(l\)$$ has gradient $$\(-\frac{2}{3}\)$$ and passes through $$\(A\)$$. A circle has centre $$\((5,11)\)$$ and radius $$\(\sqrt{52}\)$$. Show that $$\(l\)$$ is the tangent to the circle at $$\(A\)$$. .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... ....................................................................................................................................................
Exam No：9709_s21_qp_12 Year：2021 Question No：7(a)
Answer:
\(
(5-1)^{2}+(11-5)^{2}=52 \text { or } \frac{11-5}{5-1}
\)
For both circle equation and gradient, and proving line is perpendicular and stating that \(A\) lies on the circle
Alternative method for Question 7(a)
\((x-5)^{2}+(y-11)^{2}=52\) and \(y-5=-\frac{2}{3}(x-1)\)
Solving simultaneously to obtain \((y-5)^{2}=0\) or \((x-1)^{2}=0 \Rightarrow 1\) root or tangent or discriminant \(=0 \Rightarrow 1\) root or tangent
Alternative method for Question 7(a)
\(
\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{10-2 x}{2 y-22}=\frac{10-2}{10-22}
\)
Showing gradient of circle at \(\mathrm{A}\) is \(-\frac{2}{3}\)
(5-1)^{2}+(11-5)^{2}=52 \text { or } \frac{11-5}{5-1}
\)
For both circle equation and gradient, and proving line is perpendicular and stating that \(A\) lies on the circle
Alternative method for Question 7(a)
\((x-5)^{2}+(y-11)^{2}=52\) and \(y-5=-\frac{2}{3}(x-1)\)
Solving simultaneously to obtain \((y-5)^{2}=0\) or \((x-1)^{2}=0 \Rightarrow 1\) root or tangent or discriminant \(=0 \Rightarrow 1\) root or tangent
Alternative method for Question 7(a)
\(
\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{10-2 x}{2 y-22}=\frac{10-2}{10-22}
\)
Showing gradient of circle at \(\mathrm{A}\) is \(-\frac{2}{3}\)
Knowledge points:
1.3.4 use algebraic methods to solve problems involving lines and circles (Including use of elementary geometrical properties of circles, e.g. tangent perpendicular to radius, angle in a semicircle, symmetry.) (Implicit differentiation is not included.)
Solution:
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