If the equation of the common tangent at the | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let $A$ and $B$ be the centers of the two circles. Since the line $12x + 5y - 7 = 0$ is tangent to both circles at $(1, -1)$,the line connecting t

Vedclass · Accepted Answer

$(13, 4), (-11, -6)$

Vedclass · Answer

(B) Let $A$ and $B$ be the centers of the two circles. Since the line $12x + 5y - 7 = 0$ is tangent to both circles at $(1, -1)$,the line connecting the centers $AB$ is perpendicular to the tangent at $(1, -1)$. The slope of the tangent is $m_t = -\frac{12}{5}$. Therefore,the slope of the line $AB$ is $m_{AB} = -\frac{1}{m_t} = \frac{5}{12}$. Let $	an 	heta = \frac{5}{12}$,so $\cos 	heta = \frac{12}{13}$ and $\sin 	heta = \frac{5}{13}$. The centers $A$ and $B$ lie on the line $AB$ at a distance of $13$ units from the point $(1, -1)$. Using the parametric form,the coordinates are $(1 \pm 13 \cos 	heta, -1 \pm 13 \sin 	heta)$. Substituting the values: $(1 \pm 13 \cdot \frac{12}{13}, -1 \pm 13 \cdot \frac{5}{13}) = (1 \pm 12, -1 \pm 5)$. This gives the two centers as $(1 + 12, -1 + 5) = (13, 4)$ and $(1 - 12, -1 - 5) = (-11, -6)$.

If the equation of the common tangent at the point $(1, -1)$ to the two circles,each of radius $13$,is $12x + 5y - 7 = 0$,then the centers of the two circles are

Similar Questions

Let $A$ be the centre of the circle $x^{2}+y^{2}-2x-4y-20=0$. Let $B(1,7)$ and $D(4,-2)$ be two points on the circle such that tangents at $B$ and $D$ meet at $C$. The area of the quadrilateral $ABCD$ is

The equation of the normal to the circle $2x^2 + 2y^2 - 2x - 5y + 3 = 0$ at the point $(1, 1)$ is:

If $P(\frac{\pi}{4})$ and $Q(\frac{\pi}{3})$ are two points on the circle $x^2+y^2-2x-2y-1=0$,then the slope of the tangent to this circle which is parallel to the chord $PQ$ is

If the line $3x - 4y = \lambda$ touches the circle $x^2 + y^2 - 4x - 8y - 5 = 0$,then $\lambda$ is equal to

The line $(x - a)\cos \alpha + (y - b)\sin \alpha = r$ will be a tangent to the circle $(x - a)^2 + (y - b)^2 = r^2$:

Vedclass Test Series

Exam Paper Generator

Online Exam Module