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}

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