22.7 Practice Problems
- (a).
- By completing the square, find the centre and radius of the circle \(x^{2}+y^{2}-2x+6y+6=0\).
- (b).
- Hence find the equation of the tangent to the circle at \(\left (2,\sqrt {3}-3\right )\).
Show solution
Solution. (a). Group and complete the square in each variable: \[\left (x^{2}-2x\right )+\left (y^{2}+6y\right )+6=0 \implies (x-1)^{2}-1+(y+3)^{2}-9+6=0,\] so \((x-1)^{2}+(y+3)^{2}=4\): centre \((1,-3)\) and radius \(2\).
(b). First check the point is on the circle: \((2-1)^{2}+\left (\sqrt {3}-3+3\right )^{2}=1+3=4\). It is.
A tangent is perpendicular to the radius at the point of contact, so no differentiation is needed. The radius from \((1,-3)\) to \(\left (2,\sqrt {3}-3\right )\) has gradient \[m_{r}=\frac {\left (\sqrt {3}-3\right )-(-3)}{2-1}=\sqrt {3},\] so the tangent has gradient \(-\frac {1}{\sqrt {3}}\) and its equation is \[y-\left (\sqrt {3}-3\right )=-\frac {1}{\sqrt {3}}(x-2).\]
Note 22.13. Using the radius rather than implicit differentiation is worth the habit: for a circle the perpendicularity is immediate, and it avoids the algebra entirely. The two routes agree, as they must — differentiating implicitly gives \(\frac {dy}{dx}=\frac {1-x}{y+3}\), which at this point is \(\frac {-1}{\sqrt {3}}\), the same answer.
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