Finding equation of circle (A Maths) 8 mark 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.


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}



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

Hence the equation of the circle is:



Logarithm and Exponential Question: A Maths Question


Solve (4x)^{\lg 5} = (5x)^{\lg 7}


4^{\lg 5}\cdot x^{\lg 5}=5^{\lg 7}\cdot x^{\lg 7}

\displaystyle\frac{4^{\lg 5}}{5^{\lg 7}}=\frac{x^{\lg 7}}{x^{\lg 5}}=x^{\lg 7-\lg 5}

Using calculator, and leaving answers to at least 4 s.f.,


Lg both sides,

\lg 0.6763=0.1461\lg x

\lg x=\frac{\lg 0.6763}{0.1461}=-1.1626

x=10^{-1.1626}=0.0688 (3 s.f.)

Check answer (to prevent careless mistakes):

LHS=(4\times 0.0688)^{\lg 5}=0.406

RHS=(5\times 0.0688)^{\lg 7}=0.406

Since LHS=RHS, we have checked that our answer is valid.

Maths Tutor Singapore, H2 Maths, A Maths, E Maths

If you or a friend are looking for maths tuition: o level, a level, IB, IP, olympiad, GEP and any other form of mathematics you can think of

Experienced, qualified (Raffles GEP, Deans List, NUS Deans List, Olympiads etc) and most importantly patient even with the most mathematically challenged.

so if you are in need of the solution to your mathematical woes, drop me a message!

Tutor: Mr Wu