Question

# A circle passes through the point (2,3) and (4,5). If it's centre lies on the line y - 4x + 3 = 0, then its radius is equal to:a. $\sqrt 5$b. 1c. $\sqrt 2$d. 2

Verified
154.5k+ views

Hint: Concept involved here is that the centre of the circle lies on the perpendicular bisector of AB. Where A is (2,3) and B is (4,5). Let’s find the equation of the line passing through these two points and find the midpoint of AB. Then, find the bisector of AB passes through the midpoint and with elimination method, find out values of x and y to find the intersection point of two lines. Now, find out AC by distance formula.

Complete step by step answer:

Equation of the line passing through two points is $y - {y_1} = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}(x - {x_1})$

So, using above we can write the equation of line passing through (2,3) and (4,5) that is AB.

y - 3 = $\dfrac{{5 - 3}}{{4 - 2}}(x - 2)$

So, we get x - y + 1 = 0

Now mid point of AB is $\left( {\dfrac{{{{\text{x}}_1} + {{\text{x}}_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2}} \right)$

Mid point is $\left( {\dfrac{{2 + 4}}{2},\dfrac{{3 + 5}}{2}} \right) = \left( {3,4} \right)$

Now, the $\bot$ bisector of AB passes through the midpoint (3,4)

So, equation of $\bot$ bisector is

$y - 4 = - \dfrac{1}{m}\left( {x - 3} \right)$ where m is slope of AB

m = 1

So, the equation of $\bot$ bisector is x + y - 7 = 0

Now the point of intersection of two lines is y - 4x + 3 = 0 and this $\bot$ bisector is the centre of the circle.

Using elimination method we can find the intersection of the two lines.

So, the centre comes out to be C (2,5)

So, the radius of the circle is AC.

${\text{AC = }}\sqrt {{{\left( {2 - 2} \right)}^2} + {{\left( {3 - 5} \right)}^2}}$

AC = 2units

Hence (d) is the right answer.

Note: In such questions always use the concept that the centre of the circle always passes through the $\bot$ bisector of the two given passing points of the circle.