SATHEE: Circle 4 Question 11

Circle 4 Question 11

11. The equation of the circle passing through $(1, 1)$ and the points of intersection of $x^{2} + y^{2} + 13 x - 3 y = 0$ and $2 x^{2} + 2 y^{2} + 4 x - 7 y - 25 = 0$ is

(1983, 1M)

(a) $4 x^{2} + 4 y^{2} - 30 x - 10 y = 25$

(b) $4 x^{2} + 4 y^{2} + 30 x - 13 y - 25 = 0$

(d) None of the above

Show Answer

Answer:

Correct Answer: 11. (b)

Solution:

The required equation of circle is

$(x^{2} + y^{2} + 13 x - 3 y) + λ 11 x + \frac{1}{2} y + \frac{25}{2} = 0$

Its passing through $(1, 1)$ ,

$\begin{aligned} \Rightarrow & 12 + λ (24) & = 0 \\ \Rightarrow & λ & = - \frac{1}{2} \end{aligned}$

On putting in Eq. (i), we get

$\begin{aligned} x^{2} + y^{2} + 13 x - 3 y - \frac{11}{2} x - \frac{1}{4} y - \frac{25}{4} & = 0 \\ \Rightarrow & 4 x^{2} + 4 y^{2} + 52 x - 12 y - 22 x - y - 25 & = 0 \\ \Rightarrow & 4 x^{2} + 4 y^{2} + 30 x - 13 y - 25 & = 0 \end{aligned}$

Circle 4 Question 11

11. The equation of the circle passing through $(1, 1)$ and the points of intersection of $x^{2} + y^{2} + 13 x - 3 y = 0$ and $2 x^{2} + 2 y^{2} + 4 x - 7 y - 25 = 0$ is

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Circle 4 Question 11

11. The equation of the circle passing through (1,1) and the points of intersection of x2+y2+13x−3y=0 and 2x2+2y2+4x−7y−25=0 is

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Menu

11. The equation of the circle passing through $(1, 1)$ and the points of intersection of $x^{2} + y^{2} + 13 x - 3 y = 0$ and $2 x^{2} + 2 y^{2} + 4 x - 7 y - 25 = 0$ is