## Finding equation of circle (A Maths) 8 mark Question!

Question:

Find the equation of the circle which passes through $A(8,1)$ and $B(7,0)$ and has, for its tangent at $B$, the line $3x-4y-21=0$.

Solution:

Recall that the equation of a circle is $(x-a)^2+(y-b)^2=r^2$, where $(a,b)$ is the centre of the circle, and $r$ is the radius of the circle.

Substituting $A(8,1)$ into the equation, we get: $\boxed{(8-a)^2+(1-b)^2=r^2}$ — Eqn (1)

Substituting $B(7,0)$, we get: $\boxed{(7-a)^2+(0-b)^2=r^2}$ — Eqn (2)

Equating Eqn (1) and Eqn (2), we get $64-16a+a^2+1-2b+b^2=49-14a+a^2+b^2$ which reduces to $\boxed{b=8-a}$ — Eqn (3)

after simplification.

Now, we rewrite the equation of the tangent as $\displaystyle y=\frac{3}{4}x-\frac{21}{4}$ (make y the subject)

Hence, the gradient of the normal is $\displaystyle\frac{-1}{\frac{3}{4}}=-\frac{4}{3}$

Let the equation of the normal be $\displaystyle y=-\frac{4}{3}x+c$

Substitute in $B(7,0)$ we get $\displaystyle 0=-\frac{4}{3}(7)+c$

Hence $\displaystyle c=\frac{28}{3}$

Thus equation of normal is $\displaystyle \boxed{y=-\frac{4}{3}x+\frac{28}{3}}$

Since the normal will pass through the centre $(a,b)$ we have $\boxed{b=-\frac{4}{3}a+\frac{28}{3}}$ — Eqn (4)

Finally, we equate Eqn (3) and Eqn (4), $\displaystyle 8-a=-\frac{4}{3}a+\frac{28}{3}$ $\displaystyle \frac{1}{3}a=\frac{4}{3}$ $a=4$ $b=8-a=4$

Substituting back into Eqn (1), we get $r=5$

Hence the equation of the circle is: $\displaystyle\boxed{(x-4)^2+(y-4)^2=5^2}$ 1. iamvistinginstructoreric says: