Question

The circle passes through the points (-1,0) and touching the y-axis at (0,2) also passes through the point
A. $\left( { - \dfrac{3}{1},0} \right)$.
B. $\left( { - \dfrac{5}{2},2} \right)$
C. $\left( { - \dfrac{3}{2} - \dfrac{5}{2}} \right)$
D. $\left( { - 4,0} \right)$

Answer 1

Hint: Get the center of the circle with some arbitrary constants. Then solve the arbitrary constants, to find the radius. With the centre and radius we get the equation of the circle.

Complete step by step solution:
Since a tangent is always perpendicular to a circle its point of contact will be the foot of the perpendicular from the centre.
$Therefore, $ the y coordinate of the centre is also 2.
Let the center be (x,2) we know distance from center to any point of the circumference is equal and same that of radius distance from (d,2) is same from (0,2) & (-1,0) = radius (r)
$Therefore, r = \sqrt {{{\left( {2 - 2} \right)}^2} + {x^2}} $ (1)
$r = \sqrt {{{\left( {2 - 0} \right)}^2} + {{\left( {x + 1} \right)}^2}} $ (2)
Equating (1) & (2)
$x = \sqrt {4 + \left( {{x^2} + 1 + 2x} \right)} $
Squaring on both sides
$4 + 1 + 2x = 0$
$x = - \dfrac{5}{2}$
Substituting in (1)
$r = \sqrt {{r^2}} = (r) = \dfrac{5}{2}$
Equation of circle is given by ${\left( {x - {x_1}} \right)^2} + {\left( {y - {y_1}} \right)^2} = {r^2}$ where $\left( {{x_1},{y_1}} \right)$ is the center & r is the radius.
$Therefore, $ $E{q^n}$ of circle is ${\left( {x + \dfrac{5}{2}} \right)^2} + {\left( {y - 2} \right)^2} = \dfrac{{25}}{4}.$

Substituting the options in the circle equation we observe that the (-4,0) lies on the circle must satisfy the circle equation.

Note: Always write the general equation of the circle to understand the parameters that are required to be found. We can also solve this question with a tangent equation, as it is given that the y-axis is tangent to the circle.