Question

$x^2+y^2+px+3y-5=0and{x}^2+y^2+5x+py+7=0$ cuts orthogonally, then P is _______

• A. $\frac{1}{2}$
• B. 1
• C. $\frac{3}{2}$
• D. 2

Answer 1

The correct option is A

$\frac{1}{2}$

Two circles $S_1=0and{S}_2=0$ are said to be orthogonal if the tangents at their point of intersection

include a right angle.

The condition for two circles to be orthogonal is:-

$2g_1{g}_2+2f_1{f}_2=c_1+c_2$ ..............(1)

Proof:

If two circles are $x^2+y^2+2g_{1}x+2f_{1}y+c_1=0\&x^2+y^2+2g_{2}x+2f_{2}y+c_2=0$

Let center of circles being $(-g_1,-f_1)and(-g_2,-f_2)$

In right angled triangle $C_{1}P{C}_2$

$(C_1{C}_2{)}^2=(C_{1}P{)}^2+(C_{2}P{)}^2$

$(C_1{C}_2{)}^2=r_1^2+r_2^2$

$(g_1-g_2{)}^2+(f_1-f_2{)}^2=g_1^2+f_1^2-c_1+g_2^2+f_2^2-c_2$

$2g_1{g}_2+2f_1{f}_2=c_1+c_2$

comparing given circles with standard circles and substututing $g_1,g_2,f_1,f_2,c_1\&c_2$ values in equation (1)

$g_1=\frac{-P}{2},g_2=\frac{-5}{2},f_1=\frac{-3}{2},f_2=\frac{-P}{2},c_1=-5,c_2=7$

$=2\left(\frac{-p}{2}\right)\left(\frac{-5}{2}\right)+2\left(\frac{-3}{2}\right)\left(\frac{-p}{2}\right)=-5+7$

$=P\left(\frac{5}{2}\right)+P\left(\frac{3}{2}\right)=-5+7$

$4P=2$

$P=\frac{1}{2}$

Option A is correct.