A circle has centre at the point $$\(B(5,1)\)$$. The point $$\(A(-1,-2)\)$$ lies on the circle. Point $$\(C\)$$ is such that $$\(A C\)$$ is a diameter of the circle. Point $$\(D\)$$ has coordinates $$\((5,16)\)$$. Show that $$\(D C\)$$ is a tangent to the circle. .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... ....................................................................................................................................................
Exam No：9709_w20_qp_12 Year：2020 Question No：9(b)
Answer:
$C$ has coordinates $(11,4)$
$0.5$
Grad of CD $=-2$
$\left(\frac{1}{2} \times-2=-1\right)$ then states $+$ perpendicular $\rightarrow$ hence shown or tangent
Alternative method for question $9(b)$
$C$ has coordinates $(11,4)$
$0.5$
Gradient of the perpendicular is $-2$
$\rightarrow$ Equation of the perpendicular is $y-4=-2(x-11)$
Checks $D(5,16)$ or checks gradient of $C D$ and then states $D$ lies on the line or $C D$ has gradient $-2 \rightarrow$ hence shown or tangent
Alternative method for question 9 (b)
$C$ has coordinates $(11,4)$ or Gradient of $A B, B C$ or $A C=0.5$
Equation of the perpendicular is $y-4=-2(x-11)$
$(x-5)^{2}+(-2 x+26-1)^{2}=45 \rightarrow\left(x^{2}-22 x+121=0\right)$
$(x-11)^{2}=0$ or $b^{2}-4 a c=0 \rightarrow$ repeated root $\rightarrow$ hence shown or tangent
Alternative method for question 9(b)
$C$ has coordinates $(11,4)$
Finding $C D=\sqrt{180}$ and $B D=\sqrt{225}$
Checking (their BD $)^{2}-(\text { their } \mathrm{CD})^{2}$ is the same as (their $\left.\mathrm{r}\right)^{2}$
$\therefore$ Pythagoras valid $\therefore$ perpendicular $\rightarrow$ hence shown or tangent
Alternative method for question $9(b)$
$C$ has coordinates $(11,4)$
Finding vectors $\overrightarrow{A C}$ and $\overrightarrow{C D}$ or $\overrightarrow{B C}$ and $\overrightarrow{C D}$
$\left(=\left(\begin{array}{l}6 \\ 3\end{array}\right)\right.$ and $\left(\begin{array}{c}-6 \\ 12\end{array}\right)$ or $\left(\begin{array}{c}12 \\ 6\end{array}\right)$ and $\left.\left(\begin{array}{c}-6 \\ 12\end{array}\right)\right)$
Applying the scalar product to one of these pairs of vectors
Scalar product $=0$ then states $\therefore$ perpendicular $\rightarrow$ hence shown or tangent
$0.5$
Grad of CD $=-2$
$\left(\frac{1}{2} \times-2=-1\right)$ then states $+$ perpendicular $\rightarrow$ hence shown or tangent
Alternative method for question $9(b)$
$C$ has coordinates $(11,4)$
$0.5$
Gradient of the perpendicular is $-2$
$\rightarrow$ Equation of the perpendicular is $y-4=-2(x-11)$
Checks $D(5,16)$ or checks gradient of $C D$ and then states $D$ lies on the line or $C D$ has gradient $-2 \rightarrow$ hence shown or tangent
Alternative method for question 9 (b)
$C$ has coordinates $(11,4)$ or Gradient of $A B, B C$ or $A C=0.5$
Equation of the perpendicular is $y-4=-2(x-11)$
$(x-5)^{2}+(-2 x+26-1)^{2}=45 \rightarrow\left(x^{2}-22 x+121=0\right)$
$(x-11)^{2}=0$ or $b^{2}-4 a c=0 \rightarrow$ repeated root $\rightarrow$ hence shown or tangent
Alternative method for question 9(b)
$C$ has coordinates $(11,4)$
Finding $C D=\sqrt{180}$ and $B D=\sqrt{225}$
Checking (their BD $)^{2}-(\text { their } \mathrm{CD})^{2}$ is the same as (their $\left.\mathrm{r}\right)^{2}$
$\therefore$ Pythagoras valid $\therefore$ perpendicular $\rightarrow$ hence shown or tangent
Alternative method for question $9(b)$
$C$ has coordinates $(11,4)$
Finding vectors $\overrightarrow{A C}$ and $\overrightarrow{C D}$ or $\overrightarrow{B C}$ and $\overrightarrow{C D}$
$\left(=\left(\begin{array}{l}6 \\ 3\end{array}\right)\right.$ and $\left(\begin{array}{c}-6 \\ 12\end{array}\right)$ or $\left(\begin{array}{c}12 \\ 6\end{array}\right)$ and $\left.\left(\begin{array}{c}-6 \\ 12\end{array}\right)\right)$
Applying the scalar product to one of these pairs of vectors
Scalar product $=0$ then states $\therefore$ perpendicular $\rightarrow$ hence shown or tangent
Knowledge points:
1.3.2 interpret and use any of the forms in solving problems (Including calculations of distances, gradients, midpoints, points of intersection and use of the relationship between the gradients of parallel and perpendicular lines.)
Solution:
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