### Tangents to circle, cyclic

Question Sample Titled 'Tangents to circle, cyclic'

In the figure, ${C}_{{1}}$ and ${C}_{{2}}$ are two circles touching each other at point ${M}$ . ${T}{C}{A}$ and ${T}{D}{B}$ are two common tangents to the circles. ${T}{C}{A}$ touches ${C}_{{1}}$ and ${C}_{{2}}$ at ${C}$ and ${A}$ respectively. ${T}{D}{B}$ touches ${C}_{{1}}$ and ${C}_{{2}}$ at ${D}$ and ${B}$ respectively.

${T}$${A}$${C}$${D}$${B}$${M}$${C}_{{1}}$${C}_{{2}}$

 (a) Prove that ${C}{A}{B}{D}$ is a cyclic quadrilateral. (3 marks) (b) Zac claims that ${C}{B}$ must not be a diameter of circle ${C}{A}{B}{D}$ . Do you agree? Explain your answer. (2 marks)

 (a) ∵  ${T}{C}{A}$ and ${T}{D}{B}$ are common tangents to ${C}_{{1}}$ and ${C}_{{2}}$ . ∴ ${T}{C}$ $={T}{D}$ and ${T}{A}={T}{B}$. tangent properties $\angle{T}{C}{D}$ $=\angle{T}{D}{C}$ and $\angle{T}{A}{B}=\angle{T}{B}{A}$ base ∠s, isos. △ 1M  Let $\angle{C}{T}{D}={x}$ . $\angle{T}{C}{D}$ $=\angle{T}{D}{C}=\dfrac{{{180}^{\circ}-{x}}}{{2}}$ ∠ sum of △ 1M $\angle{T}{A}{B}$ $=\angle{T}{B}{A}=\dfrac{{{180}^{\circ}-{x}}}{{2}}$ ∠ sum of △ ∴  $\angle{T}{C}{B}$ $=\angle{A}{B}{D}$ 1M ∴  ${C}{A}{B}{D}$ is a cyclic quadrilateral. ext. ∠ = int. opp. ∠  (b) From (a), $\angle{C}{A}{B}$ $=\dfrac{{{180}^{\circ}-{x}}}{{2}}$  $={90}^{\circ}-\dfrac{{x}}{{2}}$  $\ne{90}^{\circ}$ 1M ∴  ${C}{B}$ is not a diameter of ${C}{A}{B}{D}$. Thus, he is agreed. 1A

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