Question

Question 14
If the centre of a circle is (2a, a-7), then find the values of a , if the circle passes through the point (11,-9) and has diameter $10\sqrt{2}$ units.

Answer 1

By given condition,

Distance between the centre C(2a , a-7) and the point P(11,-9), which lie on the circle = Radius of circle
$\therefore \text{Radius of circle}=\sqrt{(11-2a{)}^2+(-9-a+7{)}^2}...\left(i\right)\left[\begin{array}{c}\because \text{distance between the points}(x_1,y_1)and(x_2,y_2)d\\ =\sqrt{(x_2-x_1{)}^2+(x_2-y_1{)}^2}\end{array}\right]\text{Given that, length of diameter =}10\sqrt{2}\therefore \text{Length of radius}=\frac{Lengthofdiameter}{2}=\frac{10\sqrt{22}}{2}=5\sqrt{2}\text{Put this value in Eq.(i), we get}50=(11-2a{)}^2+(2+a{)}^2\Rightarrow 50=121+4a^2-44a+4+a^2+4a\Rightarrow 5a^2-40a+75=0\Rightarrow a^2-8a+15=0\Rightarrow a^2-5a-3a+15=0\text{[by splitting the middle term]}\Rightarrow a(a-5)-3(a-5)=0\Rightarrow (a-5)(a-3)=0\therefore a=3,5\text{Hence, the required values of a are 5 and 3.}$