Question

The line 4x-3y=-12 is the tangent at point A(-3,0) and the line 3x+4y=16 is the tangent at the point B(4,1) to a circle. The equation of circle is .

• A. $(x+1{)}^2+(y+3{)}^2=25$
• B. $(x+1{)}^2+(y-3{)}^2=25$
• C. $(x-1{)}^2+(y+3{)}^2=25$
• D. $(x-1{)}^2+(y-3{)}^2=25$

Answer 1

The correct option is C $(x-1{)}^2+(y+3{)}^2=25$

We know that the center of a circle will be the intersection point of normals. Let the center of the given circle be C.
the equation of the line through A perpendicular to the line 4x - 3y = -12 will be:
$y-0=\frac{-3}{4}(x+3)\Rightarrow 3x+4y+9=0...\left(i\right)$
and the equation of the line passing through B perpendicular to the line 3x+4y=16 will be:
$y-1=\frac{4}{3}(x-4)\Rightarrow 4x-3y-13=0...\left(\mathrm{ii}\right)$On solving (i) and (ii), we get:
x = 1 and y = -3
So, the coordinates of C are (1,-3).
and Radius = CA = 5
Hence, the equation of the circle will be :
$(x-1{)}^2+(y+3{)}^2=25$