Question

The equation of the circle passing through the point (–1, –3) and touching the line 4x + 3y – 12 = 0 at the point (3, 0), is

• A.
• B.
• C. 2
• D. None of these

Answer 1

The correct option is A

Let the equation be
$x^2+y^2+2gx+2fy+c=0\dots \dots \dots \dots \left(i\right)$
But it passes through (-1, -3) and (3, 0) therefore
10 -2g - 6f + c = 0 ------- (ii)
9 + 6g + c = 0 ------(iii)
Also centre is C(-g, -f).
Slope of tangents = $-\frac{4}{3}\Rightarrow$Slope of normal = $\frac{3}{4}$
$\Rightarrow \frac{f}{3+g}=\frac{3}{4}\Rightarrow 3g-4f+9=0$
Now on solving (ii), (iii) and (iv), we get
g = -1, f = $\frac{3}{2}$ and c = -3
Therefore, the equation of circle is
$x^2+y^2-2x+3y-3=0$
Trick: The points (-1, -3) and (3, 0) must satisfy the elution of circle. Circle given in (a) satisfies both the points. Also check whether it touches the line 4x + 3y - 12 = 0 or not.